The correct choice that indicates an elastic collision is: "No correct choice is available in the list."
An elastic collision is defined as a collision where kinetic energy is conserved, and the objects rebound without any loss of energy. In an elastic collision, the objects involved do not become hotter, get stuck together, or deform.
"Objects are hotter after collision": In an elastic collision, the total kinetic energy of the system remains the same before and after the collision. If the objects become hotter after the collision, it implies an increase in their internal energy, which would indicate that energy was not conserved. Therefore, an increase in temperature would suggest an inelastic collision, not an elastic one.
"Both objects get stuck together after collision": If the objects stick together and move as a single unit after the collision, it suggests that there was a loss of kinetic energy during the collision. In an elastic collision, the objects separate after the collision, maintaining their individual identities and velocities. Therefore, objects getting stuck together implies an inelastic collision, not an elastic one.
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The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 11 m/s. (a) What is the magnitude of the velocity of the projectile 1.8 s before it achieves its maximum height? (b) What is the magnitude of the velocity of the projectile 1.8 s after it achieves its maximum height? If we take x = 0 and y = 0 to be at the point of maximum height and positive x to be in the direction of the velocity there, what are the (c) x coordinate and (d) y coordinate of the projectile 1.8 s before it reaches its maximum height and the (e) x coordinate and (f) y coordinate 1.8 s after it reaches its maximum height?
(a) The magnitude of the velocity 1.8 s before reaching the maximum height is approximately 8.14 m/s. (b) The magnitude of the velocity 1.8 s after reaching the maximum height is approximately
To calculate the magnitude of the velocity of the projectile 1.8 s before it reaches its maximum height, we can use the principle of conservation of energy. At its maximum height, all the initial kinetic energy is converted to potential energy.
(a) The magnitude of the velocity at maximum height is 11 m/s, we can calculate the velocity 1.8 s before using the equation for conservation of energy:
Potential energy at maximum height = Kinetic energy 1.8 s before maximum height
mgh = (1/2)mv^2
where m is the mass of the projectile, g is the acceleration due to gravity, h is the maximum height, and v is the velocity.
Since the mass and acceleration due to gravity are constant, we can write:
h = (1/2)v^2 / g
Substituting the given values, we have:
h = (1/2)(11^2) / 9.8
h ≈ 6.04 m
Now, using the equations of motion for vertical motion:
v = u + gt
where u is the initial velocity (which is the velocity at maximum height) and g is the acceleration due to gravity.
Substituting the values:
v = 11 + (-9.8)(1.8)
v ≈ -8.14 m/s (negative sign indicates the velocity is in the opposite direction)
Therefore, the magnitude of the velocity 1.8 s before reaching the maximum height is approximately 8.14 m/s.
(b) To calculate the magnitude of the velocity 1.8 s after reaching the maximum height, we can use the same approach. The equations of motion remain the same, but the initial velocity will now be the velocity at the maximum height.
v = u + gt
v = 11 + (9.8)(1.8)
v ≈ 27.24 m/s
Therefore, the magnitude of the velocity 1.8 s after reaching the maximum height is approximately 27.24 m/s.
(c) and (d) To determine the x and y coordinates 1.8 s before reaching the maximum height, we can use the equations of motion:
x = uxt
y = uyt + (1/2)gt^2
Since the projectile is at its maximum height, the y-coordinate will be the maximum height (h) and the y-velocity (uy) will be zero. Substituting the values, we have:
x = (11)(1.8) = 19.8 m
y = 6.04 m
Therefore, the x-coordinate 1.8 s before reaching the maximum height is approximately 19.8 m and the y-coordinate is approximately 6.04 m.
(e) and (f) To calculate the x and y coordinates 1.8 s after reaching the maximum height, we can use the same equations:
x = uxt
y = uyt + (1/2)gt^2
Since the projectile is at its maximum height, the y-coordinate will remain the same (h) and the y-velocity (uy) will still be zero. Substituting the values, we have:
x = (11)(1.8) = 19.8 m
y = 6.04 m
Therefore, the x-coordinate 1.8 s after reaching the maximum height is approximately 19.8 m and the y-coordinate remains approximately 6.04 m.
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earth - 5.9742 x 1024 kg l'earth-6.3781 x 106m mmoon - 7.36 x 1022 kg moon - 1.7374 x 106m dearth to moon - 3.844 x 108 m (center to center) G 6.67428 x 10-11 N.m/kg? A 1900 kg satellite is orbitting the earth in a circular orbit with an altitude of 1400 km. 1) How much energy does it take just to get it to this altitude? Submit 2) How much kinetic energy does it have once it has reached this altitude? U Submit 3) What is the ratio of the this change in potential energy to the change in kinetic energy?
Energy it take just to get it to this altitude is 4.594 x 10¹⁰ J.
kinetic energy it have once it has reached this altitude is 4.274 x 10¹⁰ J.
The ratio of the this change in potential energy to the change in kinetic energy is 1.075.
If the final altitude of the satellite were 4800 km, the ratio is 0.270.
If the final altitude of the satellite were 3185 km, the ratio is 0.087.
(a) The potential energy (PE) that is needed to lift a satellite of mass m to a height of h is given as PE = G M m / r, where G is the universal gravitational constant (6.67428 x 10⁻¹¹ N-m²/kg²), M is the mass of the Earth, m is the mass of the satellite, and r is the distance from the center of the Earth to the satellite. The distance from the center of the Earth to the satellite is equal to the sum of the radius of the Earth (rₑ) and the altitude (h). Therefore,
r = rₑ + h
= 6.3781 x 10⁶ m + 1700 x 10³ m
= 6.5481 x 10⁶ m
Thus, PE = G M m / r
= (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / (6.5481 x 10⁶ m + 1700 x 10³ m)
= 4.594 x 10¹⁰ J.
(b) The kinetic energy (KE) of a satellite moving in a circular orbit of radius r around a planet of mass M is given as
KE = G M m / (2 r).
Thus, KE = G M m / (2 r)
= (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / [2(6.3781 x 10⁶ m + 1700 x 10³ m)]
= 4.274 x 10¹⁰ J.
(c) The ratio of the change in potential energy (ΔPE) to the change in kinetic energy (ΔKE) is given as ΔPE / ΔKE = (PE - PE₀) / (KE - KE₀), where PE₀ and KE₀ are the initial potential and kinetic energies of the satellite. Since the satellite is initially at rest,
KE₀ = 0, and
ΔKE = KE - KE₀
= KE.
Thus, ΔPE / ΔKE = (PE - PE₀) / KE
= (4.594 x 10¹⁰ J - 0 J) / 4.274 x 10¹⁰ J
= 1.075.
(d) If the final altitude of the satellite were 4800 km, then
r = rₑ + h
= 6.3781 x 10⁶ m + 4800 x 10³ m
= 6.8581 x 10⁶ m.
Substituting this value into the expression for the potential energy, we have
PE = G M m / r
= (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / (6.8581 x 10⁶ m)
= 5.747 x 10¹⁰ J.
Thus, the change in potential energy is ΔPE = PE - PE₀
= 5.747 x 10¹⁰ J - 4.594 x 10¹⁰ J
= 1.153 x 10¹⁰ J.
The change in kinetic energy is the same as before, so
ΔKE = KE = 4.274 x 10¹⁰ J.
The ratio of the change in potential energy to the change in kinetic energy is therefore
ΔPE / ΔKE = (PE - PE₀) / KE
= (1.153 x 10¹⁰ J) / (4.274 x 10¹⁰ J)
= 0.270.
(e) If the final altitude of the satellite were 3185 km, then
r = rₑ + h
= 6.3781 x 10⁶ m + 3185 x 10³ m
= 6.6971 x 10⁶ m.
Substituting this value into the expression for the potential energy, we have
PE = G M m / r
= (6.67428 x 10⁻¹¹ N-m²/kg²)(5.9742 x 10²⁴ kg)(1200 kg) / (6.6971 x 10⁶ m)
= 4.967 x 10¹⁰ J.
Thus, the change in potential energy is
ΔPE = PE - PE₀
= 4.967 x 10¹⁰ J - 4.594 x 10¹⁰ J
= 0.373 x 10¹⁰ J.
The change in kinetic energy is the same as before, so
ΔKE = KE = 4.274 x 10¹⁰ J.
The ratio of the change in potential energy to the change in kinetic energy is therefore
ΔPE / ΔKE = (PE - PE₀) / KE
= (0.373 x 10¹⁰ J) / (4.274 x 10¹⁰ J)
= 0.087.
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ᵐearth = 5.9742 x 10²⁴ kg
ʳearth = 6.3781 x 10⁶ m
ᵐmoon = 7.36 x 10²² kg
ʳmoon = 1.7374 x 10⁶ m
ᵈearth to moon = 3.844 x 10⁸ m (center to center)
G = 6.67428 x 10⁻¹¹ N-m²/kg²
A 1200 kg satellite is orbitting the earth in a circular orbit with an altitude of 1700 km.
How much energy does it take just to get it to this altitude?
____J
How much kinetic energy does it have once it has reached this altitude?
______J
What is the ratio of the this change in potential energy to the change in kinetic energy? (i.e. what is (a)/(b)?)
_______
What would this ratio be if the final altitude of the satellite were 4800 km?
_______m
What would this ratio be if the final altitude of the satellite were 3185 km?
a-e
please
An RLC series circuit has a 2.40 2 resistor, a 120 µH inductor, and a 78.0 µF capacitor. (a) Find the circuit's impedance (in 22) at 120 Hz. 10.2 X 2 (b) Find the circuit's impedance (in ) at 5.00 k
An RLC series circuit is an electrical circuit that consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series.
The answers are:
a) The impedance of the RLC series circuit at 120 Hz is 217.4 Ω.
b) The impedance of the RLC series circuit at 5.00 kHz is 37.9 Ω.
The components are connected one after the other, forming a single loop for the flow of current. The resistor (R) provides resistance to the flow of current, converting electrical energy into heat.
The impedance determines how the circuit responds to different frequencies of alternating current. At certain frequencies, the impedance may be minimal, resulting in resonance, while at other frequencies, the impedance may be high, leading to a reduction in current flow.
RLC series circuits are widely used in electronics and electrical systems for various applications, such as filtering, signal processing, and frequency response analysis.
(a) To find the impedance of the RLC series circuit at 120 Hz, we need to consider the resistive, inductive, and capacitive components.
The impedance (Z) of the circuit can be calculated using the formula:
[tex]Z = \sqrt(R^2 + (XL - XC)^2)[/tex]
where:
R = resistance = 2.40 Ω
XL = inductive reactance = 2πfL, where f is the frequency and L is the inductance
XC = capacitive reactance = 1/(2πfC), where f is the frequency and C is the capacitance
Given:
[tex]L = 120\mu H = 120 * 10^{-6} H[/tex]
[tex]C = 78.0 \mu F = 78.0 * 10^{-6} F[/tex]
f = 120 Hz
Now we can calculate the impedance:
[tex]XL = 2\pi fL = 2\pi (120 Hz)(120 * 10^{-6} H)\\XC = 1/(2\pi fC) = 1/(2\pi (120 Hz)(78.0 * 10^{-6} F))[/tex]
Calculate XL and XC:
XL = 0.0902 Ω
XC = 217.3 Ω
Substitute the values into the impedance formula:
[tex]Z = \sqrt(2.40^2 + (0.0902 - 217.3)^2)[/tex]
Calculate Z:
Z = 217.4 Ω
Therefore, the impedance of the RLC series circuit at 120 Hz is 217.4 Ω.
(b) To find the impedance of the RLC series circuit at 5.00 kHz, we follow the same steps as in part (a), but with a different frequency.
Given:
[tex]f = 5.00 kHz = 5.00 * 10^3 Hz[/tex]
Calculate XL and XC using the new frequency:
[tex]XL = 2\pi fL = 2\pi (5.00 * 10^3 Hz)(120 × 10^{-6} H)\\XC = 1/(2\pi fC) = 1/(2\pi (5.00 * 10^3 Hz)(78.0 * 10^{-6} F))[/tex]
Calculate XL and XC:
XL = 37.7 Ω
XC = 3.40 Ω
Substitute the values into the impedance formula:
[tex]Z = \sqrt(2.40^2 + (37.7 - 3.40)^2[/tex])
Calculate Z:
Z = 37.9 Ω
Therefore, the impedance of the RLC series circuit at 5.00 kHz is 37.9 Ω.
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Hence, the circuit's impedance is (2.40 - j2.64) Ω.
The given values are
Resistance, R = 2.40 Ω
Inductance, L = 120 µH
Capacitance, C = 78.0 µF
Frequency, f = 120 Hz = 0.120 kHz
Impedance formula for an RLC circuit is,
Z = R + j (XL - XC)
Here, XL is the inductive reactance, and XC is the capacitive reactance.
They are given by,
XL = 2πfL
XC = 1/2πfC
(a) At 120 Hz,
XL = 2πfL
= 2 × 3.14 × 120 × 120 × 10⁻⁶
= 90.76 ΩXC
= 1/2πfC
= 1/2 × 3.14 × 120 × 78.0 × 10⁻⁶
= 169.58 Ω
So, the impedance of the circuit is,
Z = R + j (XL - XC)
= 2.40 + j (90.76 - 169.58)
≈ 2.40 - j78.82 Ω
(b) At 5.00 kHz,
XL = 2πfL
= 2 × 3.14 × 5 × 10³ × 120 × 10⁻⁶
= 37.68 ΩXC
= 1/2πfC
= 1/2 × 3.14 × 5 × 10³ × 78.0 × 10⁻⁶
= 40.32 Ω
So, the impedance of the circuit is,
Z = R + j (XL - XC)
= 2.40 + j (37.68 - 40.32)
≈ 2.40 - j2.64 Ω
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An airplane is heading N40 ∘
W at 600 km/hr and encounters a wind of 150 km/hr from the NE. Determine the resultant ground velocity of the plane. Draw a diagram to support your solution. (Round your final answer to 2 decimal places)
The resultant ground velocity of the plane is approximately 600.37 km/hr, heading N53.49°E.
To determine the resultant ground velocity of the plane, we can use vector addition. We'll break down the velocities into their horizontal and vertical components and then add them together.
Airplane velocity (with respect to the ground) = 600 km/hr, heading N40°W
Wind velocity = 150 km/hr from the NE
Let's first convert the velocities to their horizontal (East-West) and vertical (North-South) components:
Airplane velocity:
Horizontal component = 600 km/hr * cos(40°) = 600 km/hr * cos(40°) ≈ 458.37 km/hr (towards the west)
Vertical component = 600 km/hr * sin(40°) = 600 km/hr * sin(40°) ≈ 384.57 km/hr (towards the north)
Wind velocity:
Horizontal component = 150 km/hr * cos(45°) = 150 km/hr * cos(45°) ≈ 106.07 km/hr (towards the east)
Vertical component = 150 km/hr * sin(45°) = 150 km/hr * sin(45°) ≈ 106.07 km/hr (towards the north)
Now, let's add the horizontal and vertical components separately to find the resultant ground velocity:
Horizontal component of resultant velocity = Airplane horizontal component + Wind horizontal component
Horizontal component of resultant velocity = 458.37 km/hr - 106.07 km/hr ≈ 352.30 km/hr (towards the west)
Vertical component of resultant velocity = Airplane vertical component + Wind vertical component
Vertical component of resultant velocity = 384.57 km/hr + 106.07 km/hr ≈ 490.64 km/hr (towards the north)
Using the Pythagorean theorem, we can find the magnitude of the resultant ground velocity:
Magnitude of resultant ground velocity = sqrt((Horizontal component)^2 + (Vertical component)^2)
Magnitude of resultant ground velocity = sqrt((352.30 km/hr)^2 + (490.64 km/hr)^2)
Magnitude of resultant ground velocity ≈ 600.37 km/hr
Finally, we can determine the direction of the resultant ground velocity using trigonometry:
Direction = arctan(Vertical component / Horizontal component)
Direction = arctan(490.64 km/hr / 352.30 km/hr)
Direction ≈ 53.49°
Therefore, the resultant ground velocity of the plane is approximately 600.37 km/hr, heading N53.49°E.
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A 4.00-m-long pole stands vertically in a freshwater lake having a depth of 2.25 m. The Sun is 3.5 above the horizontal. Determine the length of the pole's shadow on the bottom of the lake. Draw a careful picture, labeling the incident and refracted angle. What length of the pole lu above the waterm Need Help? Head it Watch
The length of the part of the pole above the water is 4 - 2.25 = 1.75 m and the length of the pole's shadow on the bottom of the lake is = 0.75 m.
Pole length, l = 4 m
Depth of the lake, h = 2.25 m
Height of the sun, H = 3.5 m
In triangle ABE, we can apply Snell's law of refraction:
(sin θ1) / (sin θ2) = (v1) / (v2)
Where v1 and v2 are the speeds of light in the first and second media, respectively. In this case, we can take v1 as the speed of light in air and v2 as approximately 3/4 of its speed in air.
Substituting the values:
(sin θ1) / (sin θ2) = 4 / 3
By Snell's law of refraction:
θ2 = sin^(-1)((4sin θ1) / 3)
In triangle AEF, we can apply trigonometric ratios as follows:
tan θ1 = h / AE
tan θ2 = h / EF
Substituting the value of θ2:
tan θ1 = h / AE
tan(sin^(-1)((4sin θ1) / 3)) = h / EF
Squaring both sides:
tan^2(sin^(-1)((4sin θ1) / 3)) = (h^2) / (EF^2)
sin^2(sin^(-1)((4sin θ1) / 3)) = ((h^2) / (EF^2)) * (1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3))))
cos^2(sin^(-1)((4sin θ1) / 3)) = 1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3)))
But we know that:
cos^2(sin^(-1)((4sin θ1) / 3)) = 1 - sin^2(sin^(-1)((4sin θ1) / 3))
1 - sin^2(sin^(-1)((4sin θ1) / 3)) = 1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3))))
sin^2(sin^(-1)((4sin θ1) / 3)) = 1 - (1 / (1 + tan^2(sin^(-1)((4sin θ1) / 3))))
Substituting the value of sin θ1:
sin^2(sin^(-1)((4 * (2.25 / AE)) / 3)) = 1 - (1 / (1 + tan^2(sin^(-1)((4 * (2.25 / AE)) / 3))))
Let x = EF, then:
(h^2) / (x^2) * (1 / (1 + (h / x)^2)) = 1 - (1 / (1 + (4h / (3x))^2))
(h^2) / (x^2 + h^2) = 1 / (1 + (4h / (3x))^2)
x^2 = (h^2) / (1 / (1 + (4h / (3x))^2)) - h^2
x^2 = (h^2) + ((4h / (3x))^2 * h^2) / (1 + (4h / (3x))^2)
(1 + (4h / (3x))^2) * x^2 = (h^2) + ((4h / 3)^2 * h^2)
x^2 = (h^2) / (1 + (16h^2) / (9x^2))
(1 + (16h^2) / (9x^2)) * x^2 = h^2 + ((4h / 3)^2 * h^2)
x^2 = (h^2) / 9
=> x = h / 3
Therefore, the length of the pole's shadow on the bottom of the lake is 2.25 / 3 = 0.75 m. The length of the part of the pole above the water is 4 - 2.25 = 1.75 m.
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26. The lasing energy levels of a laser are separated by 2.95 eV. What wavelength of light does it emit? A. 242 nm B. 420 nm C. 636 nm D. 844 nm 27. What happens to the conductivity of a material as the energy gap decreases? A. It increases. B. It decreases. C. It remains the same. D. It follows no general rule. 28. What is the common name for a particles? A. an electron B. a positron C. helium nuclei D. high energy photons
Answer:
26.The correct answer is C. 636 nm.
To determine the wavelength of light emitted by the laser, we can use the equation:
E = hc/λ
where E is the energy of a photon,
h is Planck's constant (approximately 6.626 x 10^-34 J·s),
c is the speed of light (approximately 3.00 x 10^8 m/s), and
λ is the wavelength of light.
The energy difference between the lasing energy levels is given as 2.95 eV.
To convert this energy to joules, we can use the conversion factor:
1 eV = 1.602 x 10^-19 J
Therefore, the energy difference can be expressed as:
E = (2.95 eV) * (1.602 x 10^-19 J/eV)
we can rearrange the equation to solve for the wavelength:
λ = hc/E
Substituting the values:
λ = (6.626 x 10^-34 J·s * 3.00 x 10^8 m/s) / [(2.95 eV) * (1.602 x 10^-19 J/eV)]
λ ≈ 636 nm
Therefore, the wavelength of light emitted by the laser is approximately 636 nm.
The correct answer is C. 636 nm.
27.The correct answer is A. It increases.
As the energy gap decreases, the conductivity of a material generally increases. This is because a smaller energy gap allows more electrons to move across the band gap and contribute to the conduction of electricity.
Therefore, the correct answer is A. It increases.
28.The correct answer is C. helium nuclei.
The common name for α particles is helium nuclei.
Therefore, the correct answer is C. helium nuclei.
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62. 56. When Sputnik I was launched by the U.S.S.R. in October 1957, American scientists wanted to know as much as possible about this new artificial satellite. If Sputnik orbited Earth once every 96 min, calculate its orbital velocity and altitude. (6.2)
The orbital velocity of Sputnik I is 7.91 x 10³ m/s and its altitude is 0.75 x 10⁶ m.
When Sputnik I was launched by the U.S.S.R. in October 1957, American scientists wanted to know as much as possible about this new artificial satellite.
If Sputnik orbited Earth once every 96 min, calculate its orbital velocity and altitude. (6.2)
The expression for the period of revolution of an artificial satellite of mass m around a celestial body of mass M is given by,
T = 2π √ (R³/GM)
where, T = Period of revolution
R = Distance of the artificial satellite from the center of the earth
G = Universal Gravitational constant
M = Mass of the earth
For Sputnik I,
Period of revolution, T = 96 minutes (convert it to seconds)
T = 96 * 60
= 5760 seconds
Universal Gravitational constant,
G = 6.67 x 10⁻¹¹ Nm²/kg²
Mass of the earth, M = 5.98 x 10²⁴ kg
The altitude of Sputnik I from the surface of the earth can be calculated as,
Altitude = R - R(earth)where,
R(earth) = radius of the earth
= 6.4 x 10⁶ m
Orbital velocity of Sputnik I
Orbital velocity of Sputnik I can be calculated as,
v = 2πR/T
Substitute the value of
T = 5760 seconds and solve for v,
v = 2πR/5760m/s
Calculate R, we have
T = 2π √ (R³/GM)5760
= 2π √ (R³/(6.67 x 10⁻¹¹ x 5.98 x 10²⁴))
Solve for R,
R = (GMT²/4π²)¹/³
= [(6.67 x 10⁻¹¹ x 5.98 x 10²⁴) x (5760)²/4π²]¹/³
= 7.15 x 10⁶ m
Therefore,
Altitude = R - R(earth)
= 7.15 x 10⁶ m - 6.4 x 10⁶ m
= 0.75 x 10⁶ m
Orbital velocity, v = 2πR/T
= (2 x 3.14 x 7.15 x 10⁶ m)/5760 sec
= 7.91 x 10³ m/s
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The aim of parts A and B of this experiment was to compare and contrast the results of different methods of charging: touching a charged object to a neutral object (charging by induction) and grounding a neutral object while it is polarized (charging by contact). This experiment also demonstrates the law of conservation of charge.
The aim of part C of this experiment was to investigate how charge is distributed on the outer surfaces of two conductors, one being conical and the other being spherical. The charge distribution inside the spherical conductor was also examined.
Question: Write a suitable lab discussion for the above introduction given.
The purpose of this experiment is to examine two different methods of charging and to compare the outcomes of each one.
To perform these comparisons, a variety of techniques were employed, including charging by induction and grounding a polarized object. Additionally, this study aims to examine the law of conservation of charge.To further our understanding of how charge is distributed on the surface of conductors, we then studied two different types of conductors: spherical and conical. In doing so, we were able to investigate the distribution of charge inside a spherical conductor.
This lab experiment allowed us to examine a variety of phenomena related to charge, including how it behaves in different situations and how it is distributed within various types of conductors. By examining the results of this study, we were able to gain new insights into the nature of electricity and how it can be harnessed in various settings.
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How long will it take for 30 grams of Rn-222 to decay to 7.5g?
Half-Life: 3.823 Days
Problem 2 (30 points) Consider a long straight wire which Carries a current of 100 A. (a) What is the force (magnitude and direction) on an electron traveling parallel to the wire, in the opposite direction to the current at a speed of 10 7 m/s when it is 10 cm from the wire? (b) Find the force on the electron under the above circumstances when it is traveling perpendicularly toward the wire.
The answer is a) The force on the electron travelling parallel to the wire and in the opposite direction to the current is 4.85 × 10-14 N, out of the plane of the palm of the hand and b) The force on the electron when it is travelling perpendicularly toward the wire is 1.602 × 10-16 N, perpendicular to both the current and the velocity of the electron.
(a) The direction of the force can be found using the right-hand rule. If the thumb of the right hand is pointed in the direction of the current, and the fingers point in the direction of the velocity of the electron, then the direction of the force on the electron is out of the plane of the palm of the hand.
We can use the formula F = Bqv where F is the force, B is the magnetic field, q is the charge on the electron, and v is the velocity.
Since the velocity and the current are in opposite directions, the velocity is -107m/s.
Using the formula F = Bqv, the force on the electron is found to be 4.85 x 10-14 N.
(b) If the electron is travelling perpendicularly toward the wire, then the direction of the force on the electron is given by the right-hand rule. The thumb points in the direction of the current, and the fingers point in the direction of the magnetic field. Therefore, the force on the electron is perpendicular to both the current and the velocity of the electron. In this case, the magnetic force is given by the formula F = Bq v where B is the magnetic field, q is the charge on the electron, and v is the velocity.
Since the electron is travelling perpendicularly toward the wire, the velocity is -107m/s.
The distance from the wire is 10 cm, which is equal to 0.1 m.
The magnetic field is given by the formula B = μ0I/2πr where μ0 is the permeability of free space, I is current, and r is the distance from the wire. Substituting the values, we get B = 2 x 10-6 T.
Using the formula F = Bqv, the force on the electron is found to be 1.602 x 10-16 N.
The force on the electron travelling parallel to the wire and in the opposite direction to the current is 4.85 × 10-14 N, out of the plane of the palm of the hand. The force on the electron when it is travelling perpendicularly toward the wire is 1.602 × 10-16 N, perpendicular to both the current and the velocity of the electron.
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A bag of suqar weighs \3.50 lbon Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration is one-sixth that on Earth?
The weight of the bag of sugar on the Moon is approximately 0.583 pounds.
To calculate the weight of the bag of sugar on the Moon, we need to consider the gravitational force acting on it.
The weight of an object is given by the formula:
Weight = Mass × Acceleration due to gravity
On Earth, the bag of sugar weighs 3.50 pounds.
To convert this weight to mass, we need to divide by the acceleration due to gravity on Earth, which is approximately 9.8 m/s^2.
So, the mass of the bag of sugar is:
Mass = Weight on Earth / Acceleration due to gravity on Earth
= 3.50 pounds / 9.8 m/s^2
Now, on the Moon, the acceleration due to gravity is one-sixth of that on Earth.
Therefore, the acceleration due to gravity on the Moon is:
Acceleration due to gravity on Moon = (1/6) × 9.8 m/s^2
To find the weight on the Moon, we use the same formula:
Weight on Moon = Mass × Acceleration due to gravity on Moon
= Mass × (1/6) × 9.8 m/s^2
Substituting the value of the mass calculated earlier:
Weight on Moon = (3.50 pounds / 9.8 m/s^2) × (1/6) × 9.8 m/s^2
Simplifying this equation,
We find that the weight of the bag of sugar on the Moon is approximately 0.583 pounds.
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Mars has a mass of 6.421 × 1023kg, and radius 3.4 × 106m. (a) Calculate the gravitational acceleration "g", atthe surface of Mars. (b) Will the gravitational potential approximation given above for Mars be accurate over a larger or smaller range of values of ∆y than that for the Earth? Justify your answer (do the math).
(a) To calculate the gravitational acceleration at the surface of Mars, we can use the formula for gravitational acceleration: g=GM/r2,
where G is the gravitational constant, M is the mass of Mars, and r is the radius of Mars.
(b) To determine if the gravitational potential approximation for Mars is accurate over a larger or smaller range of values of ∆y compared to Earth, we need to compare the values of g Mars and Earth and analyze the impact of the difference in radius.
Calculation: Given:
Mass of Mars (M) = 6.421 × 10^23 kg
Radius of Mars (r) = 3.4 × 10^6 m
Gravitational constant (G) = 6.67430 × 10^-11 m^3 kg^-1 s^-2(
a) Calculate the gravitational acceleration at the surface of Mars: g=GMr2g = r2GMg= (6.67430×10−11 m3 kg−1 s−2)×(6.421×1023 kg)(3.4×106 m)2g=(3.4×106m)2(6.67430×10−11m3kg−1s−2)×(6.421×1023kg)g ≈ 3.71 m/s2g≈3.71m/s2
(b) To compare the accuracy of the gravitational potential approximation, we need to consider the change in g(∆g) as ∆y varies. The gravitational potential approximation is accurate as long as ∆y is small enough that the change in g is negligible compared to the initial value.
Therefore, the gravitational potential approximation will be accurate over a smaller range of values of ∆y on Mars compared to Earth.
Final Answer:
(a) The gravitational acceleration at the surface of Mars is approximately 3.71 m/s^2.
(b) The gravitational potential approximation for Mars will be accurate over a smaller range of values of ∆y compared to Earth due to the smaller magnitude of Δg on Mars.
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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rads). If a particular disk is spuna 704.8 rad/s while it is being read, and then is allowed to come to rest over 0.368 seconds, what is the magnitude of the average angular acceleration of the disk? rad average angular acceleration: If the disk is 0.12 m in diameter, what is the magnitude of the tangentiat acceleratie of a point 1/3 of the way out from the center of the disk? ES tangential acceleration:
The magnitude of the average angular acceleration of the disk is 760 rad/s^2. The magnitude of the tangential acceleration of a point located 1/3 of the way out from the center of the disk is approximately 76.7 m/s^2.
To calculate the average angular acceleration, we can use the formula:
Average angular acceleration = (final angular velocity - initial angular velocity) / time
Given that the initial angular velocity is 704.8 rad/s and the time taken for the disk to come to rest is 0.368 seconds, we can substitute these values into the formula:
Average angular acceleration = (0 - 704.8 rad/s) / 0.368 s
Average angular acceleration ≈ -1913.04 rad/s^2
The negative sign indicates that the disk is decelerating.
Next, to find the tangential acceleration of a point 1/3 of the way out from the center of the disk, we need to calculate the radius of that point. Since the disk has a diameter of 0.12 m, its radius is half of that, which is 0.06 m.
We can use the formula for tangential acceleration:
Tangential acceleration = radius × angular acceleration
Substituting the values, we get:
Tangential acceleration = 0.06 m × -1913.04 rad/s^2
Tangential acceleration ≈ -114.78 m/s^2
The negative sign indicates that the tangential acceleration is in the opposite direction to the motion of the point.
To obtain the magnitude of the tangential acceleration, we disregard the negative sign, resulting in approximately 114.78 m/s^2.
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The magnitude of the average angular acceleration of the disk is 760 rad/s^2. The magnitude of the tangential acceleration of a point located 1/3 of the way out from the center of the disk is approximately 76.7 m/s^2.
To calculate the average angular acceleration, we can use the formula:
Average angular acceleration = (final angular velocity - initial angular velocity) / time
Given that the initial angular velocity is 704.8 rad/s and the time taken for the disk to come to rest is 0.368 seconds, we can substitute these values into the formula:
Average angular acceleration = (0 - 704.8 rad/s) / 0.368 s
Average angular acceleration ≈ -1913.04 rad/s^2
The negative sign indicates that the disk is decelerating.
Next, to find the tangential acceleration of a point 1/3 of the way out from the center of the disk, we need to calculate the radius of that point. Since the disk has a diameter of 0.12 m, its radius is half of that, which is 0.06 m.
We can use the formula for tangential acceleration:
Tangential acceleration = radius × angular acceleration
Substituting the values, we get:
Tangential acceleration = 0.06 m × -1913.04 rad/s^2
Tangential acceleration ≈ -114.78 m/s^2
The negative sign indicates that the tangential acceleration is in the opposite direction to the motion of the point.
To obtain the magnitude of the tangential acceleration, we disregard the negative sign, resulting in approximately 114.78 m/s^2.
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Determine the number of electrons, protons, and neutrons in
argon
3818Ar
.
HINT
(a)
electrons
(b)
protons
(c)
neutrons
The number of electrons in Argon is 18, the number of protons is 18, and the number of neutrons is 20.
Now, let's proceed to the second part of the question. Here's how to determine the number of electrons, protons, and neutrons in Argon 38 :18 Ar :Since the atomic number of Argon is 18, it has 18 protons in its nucleus, which is also equal to its atomic number.
Since Argon is neutral, it has 18 electrons orbiting around its nucleus.In order to determine the number of neutrons, we have to subtract the number of protons from the atomic mass. In this case, the atomic mass of Argon is 38.
Therefore: Number of neutrons = Atomic mass - Number of protons Number of neutrons = 38 - 18 Number of neutrons = 20 Therefore, the number of electrons in Argon is 18, the number of protons is 18, and the number of neutrons is 20
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A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.40 cm wide and 11.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm
thick.
What is the maximum charge that can be stored in this capacitor?
We can calculate the capacitance of the capacitor:
C = ε₀(A/d) = (8.85 x [tex]10^{-12}[/tex] F/m) × (0.374 m² / 0.0000225 m)
≈ 1.467 x [tex]10^{-9}[/tex] F.
To find the maximum charge that can be stored in the capacitor, we need to use the formula for the capacitance of a parallel-plate capacitor:
C = ε₀(A/d)
Where:
- C is the capacitance.
- ε₀ is the vacuum permittivity, approximately equal to 8.85 x 10^(-12) F/m.
- A is the area of the plates.
- d is the separation between the plates.
Given:
- The width of each aluminum-foil sheet is 3.40 cm = 0.034 m.
- The length of each aluminum-foil sheet is 11.0 m.
- The mica strip has the same width and length.
- The thickness of the mica strip is 0.0225 mm = 0.0000225 m.
First, let's calculate the area of each plate:
A = width × length
= 0.034 m × 11.0 m
= 0.374 m²
Determine the effective separation between the plates.
d = thickness of mica + thickness of air gap
= 0.0000225 m + 0 (since air gap is negligible)
= 0.0000225 m
Now, we can calculate the capacitance of the capacitor:
C = ε₀(A/d) = (8.85 x [tex]10^{-12}[/tex] F/m) × (0.374 m² / 0.0000225 m)
≈ 1.467 x[tex]10^{-9}[/tex] F
Finally, the maximum charge that can be stored in the capacitor is given by the equation:
Q = C × V
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The maximum charge that can be stored in this capacitor is 6.46 x 10^-5 C.
The maximum charge that can be stored in the parallel-plate capacitor is 6.46 x 10^-5 C. Capacitance is the ability of an object to store an electric charge, and it is determined by the size, shape, and distance between the plates. Here, a parallel-plate capacitor is made from two aluminum-foil sheets, each 3.40 cm wide and 11.0 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick.
The capacitance of a parallel plate capacitor is given by;C=εA/d,where ε is the permittivity of free space, A is the area of each plate, and d is the distance between the plates.ε = 8.85 x 10^-12 F/m is the permittivity of free space A = (3.40 x 10^-2 m) x (11.0 m) = 0.374 m^2 is the area of each plated = 0.0225 x 10^-3 m is the distance between the plates
Therefore, the capacitance is;C=εA/d = 8.85 x 10^-12 x 0.374/0.0225 x 10^-3 = 1.47 x 10^-8 FThe maximum charge that can be stored in a capacitor is given by;Q=CV, where Q is the maximum charge, C is the capacitance, and V is the voltage applied across the capacitor.
To find the maximum charge, we can use the voltage equation,V=Ed/d = εE/d,where E is the electric field between the plates and d is the distance between the plates. Since the electric field is uniform, we have;E=V/d = εV/d^2Substituting the expression for the electric field into the capacitance equation, we have;C=εA/d = εA/V/ESimplifying for the voltage, we have;V=Q/CSubstituting the expression for the electric field into the voltage equation, we have;Q = CV = εAV/dThe maximum charge that can be stored in this capacitor is thus;Q = εAV/d = 8.85 x 10^-12 x 0.374/0.0225 x 10^-3 = 6.46 x 10^-5 C
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5) A beaker contains 2 grams of ice at a temperature of -10°C. The mass of the beaker may be ignored. Heat is supplied to the beaker at a constant rate of 2200J/minute. The specific heat of ice is 2100 J/kgk and the heat of fusion for ice is 334 x103 J/kg. How much time passes before the ice starts to melt? (8 pts)
The time it takes for the ice to start melting is approximately 8.22 minutes.
To calculate the time before the ice starts to melt, we need to consider the heat required to raise the temperature of the ice from -10°C to its melting point (0°C) and the heat of fusion required to convert the ice at 0°C to water at the same temperature.
First, we calculate the heat required to raise the temperature of 2 grams of ice from -10°C to 0°C using the specific heat formula Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat, and ΔT is the change in temperature. Substituting the given values, we get Q1 = 2 g * 2100 J/kg°C * (0°C - (-10°C)) = 42000 J.
Next, we calculate the heat of fusion required to convert the ice to water at 0°C using the formula Q = m * Hf, where Q is the heat, m is the mass, and Hf is the heat of fusion. Substituting the given values, we get Q2 = 2 g * 334 x 10³ J/kg = 668000 J.
Now, we sum up the heat required for temperature rise and the heat of fusion: Q_total = Q1 + Q2 = 42000 J + 668000 J = 710000 J.
Finally, we divide the total heat by the heat supplied per minute to obtain the time: t = Q_total / (2200 J/minute) ≈ 322.73 minutes ≈ 8.22 minutes.
Therefore, it takes approximately 8.22 minutes for the ice to start melting when heat is supplied at a constant rate of 2200 J/minute.
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Enter only the last answer c) into moodle.
A solid sphere of mass M and radius R rolls without slipping to the right with a linear speed of v
a) Find a simplified algebraic expression using symbols only for the tolal kinetic energy Kior of the ball in terms of M and R
b) IfM = 7.5 kg. R = 10,8 cm and v = 4.5 m/s find the moment of inertia of the bail.
c) Plug in the numbers from part b) into your formula from part a) to get the value of the total kinetic energy
The total kinetic energy of the rolling ball, taking into account both its translational and rotational kinetic energy, is approximately 100.356 Joules. This is calculated by considering the mass, linear speed, radius, moment of inertia, and angular velocity of the ball.
a) The total kinetic energy of the rolling ball can be expressed as the sum of its translational kinetic energy and rotational kinetic energy.
The translational kinetic energy (Kt) is given by the formula: Kt = 0.5 * M * v^2, where M is the mass of the ball and v is its linear speed.
The rotational kinetic energy (Kr) is given by the formula: Kr = 0.5 * I * ω^2, where I is the moment of inertia of the ball and ω is its angular velocity.
Since the ball is rolling without slipping, the linear speed v is related to the angular velocity ω by the equation: v = R * ω, where R is the radius of the ball.
Therefore, the total kinetic energy (Kior) of the ball can be expressed as: Kior = Kt + Kr = 0.5 * M * v^2 + 0.5 * I * (v/R)^2.
b) To find the moment of inertia (I) of the ball, we can rearrange the equation for ω in terms of v and R: ω = v / R.
Substituting the values, we have: ω = 4.5 m/s / 0.108 m = 41.67 rad/s.
The moment of inertia (I) can be calculated using the equation: I = (2/5) * M * R^2.
Substituting the values, we have: I = (2/5) * 7.5 kg * (0.108 m)^2 = 0.08712 kg·m².
c) Plugging in the values from part b) into the formula from part a) for the total kinetic energy (Kior):
Kior = 0.5 * M * v^2 + 0.5 * I * (v/R)^2
= 0.5 * 7.5 kg * (4.5 m/s)^2 + 0.5 * 0.08712 kg·m² * (4.5 m/s / 0.108 m)^2
= 91.125 J + 9.231 J
= 100.356 J.
Therefore, the total kinetic energy of the ball, with the given values, is approximately 100.356 Joules.
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A 0.5-cm tall object is placed 1 cm in front of a 2-сm focal length diverging (concave) thin lens. A person looks through the lens and sees an image. Using either ray tracing techniques or the thin lens formula, determine whether the image is a) real or virtual; b) upright or inverted; c) How far from the lens is the image located; d) How magnified or how tall is the image.
The image height is 1/3 cm and the magnification is 2/3.
Given data:Height of object, h = 0.5 cm
Focal length, f = -2 cm Object distance, u = -1 cm
The sign convention used here is that distances to the left of the lens are negative, while distances to the right are positive.
1) Determine whether the image is real or virtualThe focal length of the concave lens is negative, which indicates that it is a diverging lens. A diverging lens always forms a virtual image for any location of the object.
Therefore, the image is virtual.
2) Determine whether the image is upright or invertedThe height of the object is positive and the image height is negative. Thus, the image is inverted.
3) From the thin lens formula, we can calculate the image distance as follows:1/f = 1/v - 1/u1/-2 = 1/v - 1/-1v = 2/3 cmThe image is located 2/3 cm behind the lens.
4) The magnification is given by the following equation:m = (-image height) / (object height)h′ = m * hIn this example, the object height and the image height are both given in centimeters.
Therefore, we do not need to convert the units.
m = -v/u
= -(2/3) / (-1)
= 2/3h′
= (2/3) * (0.5)
= 1/3 cm
Therefore, the image height is 1/3 cm and the magnification is 2/3.
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PHY 103 (General Physics II) Home-Work One Due Date: May, 20 2022 Question 1. A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude that is directed at from the plane of the sheet (Fig.1). Find the magnitude of the electric flux through the sheet 2007 0.400 m -0.600 m Figure. I
The given problem describes the situation where a flat sheet that is in the form of a rectangle with sides of lengths 0.400 m and 0.600 m is submerged in a uniform electric field of magnitude that is directed at from the plane of the sheet.
The electric flux φ is given by the formula:φ = E . A . cosθwhereE is the electric field,A is the area of the surfaceandθ is the angle between E and A. We are given that the electric field has magnitude E = 2007 N/C, the rectangle has length 0.6 m and width 0.4 m, so the area of the sheet is A = (0.6 m) (0.4 m) = 0.24 m². Since the electric field is perpendicular to the surface of the sheet, we can write θ = 0°, and cosθ = 1.Using these values in the formula,φ = E . A . cosθ= (2007 N/C) (0.24 m²) (1)= 482.16 N m²/C
Answer: Therefore, the magnitude of the electric flux through the sheet is 482.16 N m²/C.
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Current Attempt in Progress A coil with an inductance of 2.6 H and a resistance of 9.412 is suddenly connected to an ideal battery with ε = 87 V. At 0.12 after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy i appearing in the resistance, and (c) energy is being delivered by the battery? (a) Number i Units <> (b) Number i Units (c) Number Po i Units
(a) The rate at which energy is being stored in the magnetic field can be calculated using the formula P = 0.5 * L * (di/dt)^2, where P is the power, L is the inductance, and di/dt is the rate of change of current.
Given that L = 2.6 H and di/dt = 0.12 A/s, substituting these values into the formula gives P = 0.5 * 2.6 * (0.12)^2 = 6.7856 W.
(b) The rate at which thermal energy is appearing in the resistance can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. At 0.12 s, the current flowing through the coil is the same as the current delivered by the battery, which is given by ε / R = 87 V / 9.412 Ω = 9.2407 A. Substituting these values into the formula gives P = (9.2407)^2 * 9.412 = 3.1557 W.
(c) The rate at which energy is being delivered by the battery is equal to the power delivered, which can be calculated using the formula P = ε * I, where P is the power, ε is the battery's electromotive force, and I is the current flowing through the coil. Substituting the given values ε = 87 V and I = 9.2407 A into the formula gives P = 87 * 9.2407 = 56.6143 W.
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1. Is the following statement true or false?
One Volt = 1 Amp/sec
2. Let there be three point charges Q1, Q2, Q3 in space arranged in an equilateral triangle formation with sides of 3mm. What is the voltage at the center of the triangle if Q1 = Q2 = 5 µC and Q3 = -7 μC ?
5.54x10^7 V
9x10^6 V
1.6x10^7 V
5.1x10^7 V
3. You shoot an electron into a capacitor as shown. What happens to the electron?
It curves down.
It curves left.
It curves up.
It keeps going straight.
The voltage at the center of the triangle is 5.54x10^7 V.
1. The "One Volt = 1 Amp/sec" is false. Voltage and current are two different quantities. Voltage is the difference in electrical potential energy between two points, while current is the rate of flow of electric charge. The unit for voltage is the volt (V), while the unit for current is the ampere (A).
2. The voltage at the center of the triangle is 5.54x10^7 V.
3. The electron will curve down.
Here are the solutions:
1. Voltage is defined as the potential difference between two points in an electric circuit. Current is defined as the rate of flow of electric charge. The unit for voltage is the volt (V), while the unit for current is the ampere (A). One volt is not equal to one amp per second.
2. The voltage at the center of the triangle can be calculated using the following formula:
V = kQ/r`
where:
* V is the voltage in volts
* k is the Coulomb constant (8.988x10^9 N⋅m^2/C^2)
* Q is the total charge in coulombs
* r is the distance between the charges in meters
In this case, the total charge is Q = 5 μC + 5 μC - 7 μC = 3 μC. The distance between the charges is r = 3 mm = 0.003 m. Plugging in these values, we get:
V = 8.988x10^9 N⋅m^2/C^2 * 3 μC / 0.003 m = 5.54x10^7 V
Therefore, the voltage at the center of the triangle is 5.54x10^7 V.
3. When an electron is shot into a capacitor, it will be attracted to the positive plate of the capacitor. The electron will curve down because the positive plate is below the electron. The electron will continue to move until it reaches the positive plate.
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Jane han conducted a virtual xperiment using a PHET simulation and completed associated lab assignment in the simulation, there was a box on the on the floor Jane appred horizontal forces on the box and measured its acceleration She recorded the mass of the box, applied force and measured acceleration values in a datatable. Then she calculated the acceleration of the box with the used mass and force. She compared the calculated value to the measured value. Which physios concept she practiced in this experiment? Free Fall Newtons Law of Motion Conservation of Energy Simple Harmonic Motion Ideal Gas Law
The physics concept that Jane practiced in this experiment is Newton's Law of Motion.
Newton's Laws of Motion describe the relationship between the motion of an object and the forces acting upon it.
In the experiment, Jane applied horizontal forces to the box on the floor and measured its acceleration.
By recording the mass of the box and the applied force, she calculated the acceleration of the box using Newton's second law, which states that the acceleration of an object is directly proportional to the net force applied to it and inversely proportional to its mass (F = ma).
After calculating the expected acceleration based on the applied force and mass, Jane compared it to the measured acceleration value.
This comparison allows her to verify whether the measured acceleration aligns with the calculated value, thereby testing the principles of Newton's Laws of Motion.
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Question 5 (1 point) A 0.02 C charge with a mass of 85.0 g is moving fast creating a magnetic field of 0.02 u T at a point Z which is 0.01 mm away from the charge. At point Z, which field, due to the
The 0.02 C charge, which has a mass of 85.0 g and is travelling quickly, produces a magnetic field of 0.02 T at point Z.
The field at point Z, due to the 0.02 C charge with a mass of 85.0 g moving fast, can be found using the formula below:
The magnetic field due to a charge in motion can be calculated using the following formula:
B = μ₀ × q × v × sin(θ) / (4πr²), where:
B is the magnetic field
q is the charge
v is the velocity
θ is the angle between the velocity and the line connecting the point of interest to the moving charge
μ₀ is the permeability of free space, which is a constant equal to 4π × 10⁻⁷ T m A⁻¹r is the distance between the point of interest and the moving charge
Given values are
q = 0.02 C
v = unknownθ = 90° (since it is moving perpendicular to the direction to the point Z)
r = 0.01 mm = 0.01 × 10⁻³ m = 10⁻⁵ m
Using the formula, B = μ₀ × q × v × sin(θ) / (4πr²)
Substituting the given values, B = (4π × 10⁻⁷ T m A⁻¹) × (0.02 C) × v × sin(90°) / (4π(10⁻⁵ m)²)
Simplifying, B = (2 × 10⁻⁵) v T where T is the Tesla or Weber per square meter
Thus, the magnetic field at point Z due to the 0.02 C charge with a mass of 85.0 g moving fast is 0.02 μT.
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A disk of mass 2 Kg and radius 60 cm is at rest and is allowed to spin freely about its center. A force of 50 N acts tangent to the edge of the wheel during 12 seconds. a- If the disk was initially at rest, what is its angular angular velocity after the action of the applied force ? b- Use the Work - Energy Theorem to calculate the angular displacement.
Given the following information: Mass of disk (m) = 2 Kg.
The radius of the disk (r) = 60 cm
Force applied (F) = 50 N
Time (t) = 12 seconds
Initial angular velocity (ωi) = 0
Find out the final angular velocity (ωf) and angular displacement (θ) of the disk.
a) The torque produced by the force is given as: T = F × r
where, T = torque, F = force, and r = radius of the disk
T = 50 N × 60 cm = 3000 Ncm
The angular acceleration (α) produced by the torque is given as:
α = T / I where, I = moment of inertia of the disk.
I = (1/2) × m × r² = (1/2) × 2 kg × (60 cm)² = 0.36 kgm²α = 3000 Ncm / 0.36 kgm² = 8333.33 rad/s².
The final angular velocity (ωf) of the disk is given as:
ωf = ωi + α × t
because the disk was initially at rest,
ωi = 0ωf = 0 + 8333.33 rad/s² × 12 sωf = 100000 rad/s.
Thus, the angular velocity of the disk is 100000 rad/s.
b)The work done (W) by the force is given as W = F × d
where d = distance traveled by the point of application of the force along the circumference of the disk
d = 2πr = 2 × 3.14 × 60 cm = 376.8 cm = 3.768 mW = 50 N × 3.768 m = 188.4 J.
The kinetic energy (Kf) of the disk after 12 seconds is given as:
Kf = (1/2) × I × ωf²Kf = (1/2) × 0.36 kgm² × (100000 rad/s)²Kf = 1.8 × 10¹² J
By the Work-Energy Theorem, we have:Kf - Ki = W
where, Ki = initial kinetic energy of the disk
Ki = (1/2) × I × ωi² = 0
Rearrange the above equation to find out the angular displacement (θ) of the disk.
θ = (Kf - Ki) / Wθ = Kf / Wθ = 1.8 × 10¹² J / 188.4 Jθ = 9.54 × 10⁹ rad.
Thus, the angular displacement of the disk is 9.54 × 10⁹ rad.
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Determine for each of the following statements whether it is correct or incorrect.
The Flectric Field at a point is numericallv egua to the force that an electron placed at the point would reel.
The magnitude of the force between two charges can be either positive or negative depending on the arrangement of the charges.
A positive charge placed at the center or a negatvely charged uniform spherical smells not in equilibrum
The Electric Field at a point due to a charge distribution is a scalar
v A charge placed at the center of a square which has 4 equal charges at its 4 corners is in equilibrium
So incorrect statements are: (1), (4)
and the correct statements are: (2), (3), (5).
Let's evaluate each statement:
1. The electric field at a point is numerically equal to the force that an electron placed at the point would feel.
- Incorrect. The electric field at a point is a measure of the force per unit charge experienced by a positive test charge placed at that point. Since an electron has a negative charge, the force it experiences would be in the opposite direction to the electric field.
2. The magnitude of the force between two charges can be either positive or negative depending on the arrangement of the charges.
- Correct. The magnitude of the force between two charges depends on their magnitudes and the distance between them, as determined by Coulomb's law. The force can be attractive (negative) if the charges have opposite signs or repulsive (positive) if the charges have the same sign.
3. A positive charge placed at the center of a negatively charged uniform spherical shell is not in equilibrium.
- Correct. If the spherical shell has a uniform negative charge distribution, it will create an electric field pointing inward towards the center. Placing a positive charge at the center would experience a repulsive force due to the electric field, indicating that the charge is not in equilibrium.
4. The electric field at a point due to a charge distribution is a scalar.
- Incorrect. The electric field at a point due to a charge distribution is a vector quantity. It has both magnitude and direction. The direction of the electric field is the direction in which a positive test charge would experience a force. The magnitude of the electric field depends on the charge distribution and the distance from the point of interest.
5. A charge placed at the center of a square which has four equal charges at its four corners is in equilibrium.
- Correct. If the charges at the four corners of the square are equal in magnitude and have opposite signs (e.g., two positive charges and two negative charges), the forces between the center charge and each corner charge will cancel out, resulting in a net force of zero. In this case, the charge at the center of the square is in equilibrium.
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1) You are watering a garden using a garden hose connected to a large open tank of water. The garden hose has a circular cross-section with a diameter of 1.4 cm, and has a nozzle attachment at its end with a diameter of 0.80 cm. What is the gauge pressure at point A in the garden hose? (Ignore viscosity for this question.)
The gauge pressure at point A in the garden hose can be calculated as follows:The gauge pressure is the difference between the absolute pressure in the hose and atmospheric pressure.
The formula to calculate absolute pressure is given by;P = ρgh + P₀Where:P is the absolute pressureρ is the density of the liquid (water in this case)g is the acceleration due to gravity h is the height of the water column above the point A.
P₀ is the atmospheric pressure. Its value is usually 101325 Pa.The height of the water column above point A is equal to the height of the water level in the tank minus the length of the hose, which is 1 meter.
Let's assume that the tank is filled to a height of 2 meters above point A.
the height of the water column above point A is given by; h = 2 m - 1 m = 1 m
The density of water is 1000 kg/m³.
A.P = ρgh + P₀P
= (1000 kg/m³)(9.81 m/s²)(1 m) + 101325 PaP
= 11025 Pa
The absolute pressure at point A is 11025 Pa.
Gauge pressure = Absolute pressure - Atmospheric pressureGauge pressure
= 11025 Pa - 101325 PaGauge pressure
= -90299 Pa
Since the gauge pressure is negative, this means that the pressure at point A is below atmospheric pressure.
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A 240−km-lang high-voitage transmission line 2.0 cm in diameter carries a steady current of 1,190 A, If the conductor is copper with a free charge density of 8.5×10 2h electro per cuble meter, how many yoars does it take one electron to travel the full length of the cable? (use 3.156×10 ^7 for the number of seconds in a year)
The time it takes one electron to travel the full length of the cable is 27.1 years.
Here's how I calculated it:
Given:
* Length of cable = 240 km = 240000 m
* Current = 1190 A
* Free charge density = 8.5 × 10^28 electrons/m^3
* Number of seconds in a year = 3.156 × 10^7 s
To find:
* Time for one electron to travel the full length of the cable (t)
1. Calculate the number of electrons in the cable:
N = nV = (8.5 × 10^28 electrons/m^3)(240000 m)^3 = 5.76 × 10^51 electrons
2. Calculate the time it takes one electron to travel the full length of the cable:
t = L/v = (240000 m) / (1190 A)(1.60 × 10^-19 C/A)(5.76 × 10^51 electrons) = 8.55 × 10^8 s = 27.1 year.
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a
camera is equipped with a lens with a focal length of 34cm. when an
object 1.1m (110cm) away is being photographed, what is the
magnification?
The magnification of the object being photographed is approximately -0.2361.
The magnification (m) of an object being photographed by a camera with a lens can be calculated using the formula:
m = -v/u
Where:
m is the magnification
v is the image distance
u is the object distance
Given:
Focal length of the lens (f) = 34 cm
Object distance (u) = 110 cm
To find the image distance (v), we can use the lens formula:
1/f = 1/v - 1/u
Substituting the known values:
1/34 = 1/v - 1/110
Simplifying the equation:
1/v = 1/34 + 1/110
Calculating this expression:
1/v = (110 + 34) / (34 × 110)
1/v = 144 / 3740
v = 3740 / 144
v ≈ 25.9722 cm
Now, we can calculate the magnification using the image distance and object distance:
m = -v/u
m = -25.9722 cm / 110 cm
m ≈ -0.2361
Therefore, the magnification of the object being photographed is approximately -0.2361.
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In a charge-to-mass experiment, it is found that a certain particle travelling at 7.0x 106 m/s is deflected in a circular arc of radius 43 cm by a magnetic field of 1.0×10− 4 T. The charge-to-mass ratio for this particle, expressed in scientific notation, is a.b ×10cdC/kg. The values of a,b,c and d are and (Record your answer in the numerical-response section below.) Your answer:
In a charge-to-mass experiment, a certain particle traveling at 7.0x10^6 m/s is deflected in a circular arc of radius 43 cm by a magnetic field of 1.0x10^-4 T.
We can determine the charge-to-mass ratio for this particle by using the equation for the centripetal force.The centripetal force acting on a charged particle moving in a magnetic field is given by the equation F = (q * v * B) / r, where q is the charge of the particle, v is its velocity, B is the magnetic field, and r is the radius of the circular path.
In this case, we have the values for v, B, and r. By rearranging the equation, we can solve for the charge-to-mass ratio (q/m):
(q/m) = (F * r) / (v * B)
Substituting the given values into the equation, we can calculate the charge-to-mass ratio.
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A 1.9 m -long string is fixed at both ends and tightened until
the wave speed is 40 m/s. What is the frequency of the standing
wave Express your answer in hertz.
The frequency of the standing wave is calculated as 10.53 Hz. The formula for frequency of the wave can be calculated by the formula: frequency = velocity / wavelength.
A 1.9 m -long string is fixed at both ends and tightened until the wave speed is 40 m/s. The velocity of the wave is given as 40 m/s and the length of the string is given as 1.9m.
The frequency of the wave can be calculated by the formula: frequency = velocity / wavelength where v is the velocity of the wave, λ is the wavelength of the wave, f is the frequency of the wave
We can calculate the wavelength of the wave using the formula given below: wavelength (λ) = 2L/n where L is the length of the string n is the harmonic number
Let's substitute the given values in the above formulas and calculate the frequency of the standing wave: wavelength (λ) = 2L/n= 2 x 1.9/1= 3.8 m
The frequency of the wave can be calculated by the formula given below: f = v/λ= 40/3.8≈ 10.53 Hz
Therefore, the frequency of the standing wave is 10.53 Hz.
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