The final velocity of the airplane is 10 m/s. This means the airplane will be moving at a speed of 10 meters per second after 40 seconds when it has decelerated from its initial velocity of 90 meters per second.
Due to the negative acceleration and velocity acting in opposite directions, it means the airplane is slowing down or decelerating.
The formula for finding the final velocity is given as:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Substitute the given values into the formula:
v = 90 + (-2.0 × 40)
v = 90 - 80
v = 10 m/s
Therefore, the final velocity of the airplane is 10 m/s. This means the airplane will be moving at a speed of 10 meters per second after 40 seconds when it has decelerated from its initial velocity of 90 meters per second.
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An RL circuit is composed of a 12 V battery, a 6.0 H inductor and a 0.050 Ohm resistor. The switch is closed at t=0 The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V. The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is zero. The time constant is 2.0 minutes and after the switch has been closed a long time the voltage across the inductor is zero
The time constant is 1.2 minutes and after the switch has been closed a long time the voltage across the inductor is 12 V.
The RL circuit described has a time constant of 1.2 minutes, and after the switch has been closed for a long time, the voltage across the inductor is 12 V.
The time constant (τ) of an RL circuit is determined by the product of the resistance (R) and the inductance (L) and is given by the formula τ = L/R. In this case, the time constant is 1.2 minutes.
When the switch is closed, current begins to flow through the circuit. As time progresses, the current increases and approaches its maximum value, which is determined by the battery voltage and the circuit's total resistance.
In an RL circuit, the voltage across the inductor (V_L) can be calculated using the formula V_L = V_0 * (1 - e^(-t/τ)), where V_0 is the initial voltage across the inductor, t is the time, and e is the base of the natural logarithm.
Given that the voltage across the inductor after a long time is 12 V, we can set V_L equal to 12 V and solve for t to determine the time it takes for the voltage to reach this value. The equation becomes 12 = 12 * (1 - e^(-t/τ)).
By solving this equation, we find that t is equal to approximately 3.57 minutes. Therefore, after the switch has been closed for a long time, the voltage across the inductor in this RL circuit reaches 12 V after approximately 3.57 minutes.
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The primary winding of a transformer is connected to a battery, a resistor, and a switch. The secondary winding is connected to an ammeter. When the switch is closed, the ammeter shows?
A. zero current
B. a non-zero current for a brief instant
C. a constant current
When the switch is closed, the ammeter will show a non-zero current for a brief instant.
When the switch is closed, it completes the circuit and allows current to flow through the primary winding of the transformer. This current induces a changing magnetic field in the core of the transformer, which in turn induces a current in the secondary winding. However, initially, there is no current flowing through the secondary winding because it takes a short moment for the induced current to build up. Therefore, the ammeter will briefly show a non-zero current before it settles to a constant value.
Option B is the correct answer: "a non-zero current for a brief instant."
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A police car with a mass of 1800 kg is headed west at 60 km/h when it has an inelastic collision with a southbound 4500 kg ambulance. The wreckage ended up travelling at 41 km/h at 65° south of west.
What was the initial velocity and direction of the ambulance? Show your vector diagram. Be sure to label your diagram and indicate direction
Is the collision above an elastic or inelastic collision? How do you know?
The collision is an inelastic collision.This can be known because in an inelastic collision, the objects stick together and move with a common velocity after the collision.
The initial velocity and direction of the ambulance:The initial velocity and direction of the ambulance can be calculated using the conservation of momentum principle which states that the total momentum before a collision is equal to the total momentum after the collision.
A police car of 1800 kg is heading west at 60 km/h and a southbound ambulance of 4500 kg has an unknown initial velocity.
Let the initial velocity of the ambulance be u m/s at angle θ with respect to the horizontal such that:u cos θ is the horizontal component of the initial velocity.u sin θ is the vertical component of the initial velocity.
Momentum before collision = Momentum after collision
Thus:1800(60) + 0 = 1800v + 4500v cos 65° + 4500v sin 65°1800v = 108000 – 34891.924v = 57.77 km/h
Let the angle the wreckage makes with the west direction be θ2. Using vector addition,The horizontal component of the wreckage velocity = v cos 65°
The vertical component of the wreckage velocity = v sin 65°
The magnitude of the wreckage velocity is 41 km/h.
Then:tanθ2 = (v sin 65°) / (v cos 65°)θ2 = 50.59° south of west
Thus the initial velocity and direction of the ambulance are 57.77 km/h at 50.59° south of west.
Therefore the collision above is an inelastic collision. This can be known because in an inelastic collision, the objects stick together and move with a common velocity after the collision. The wreckage continued to move together as a single entity after the collision.
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\begin{tabular}{ccc} \hline & 00910.0 points & Find the equivalent resistance Req between \end{tabular} Four resistors are connected as shown in the points A and B of the resistor network. figure. 1. RAB=11Ω 2. RAB=12Ω 3. RAB=16Ω 4. RAB=17Ω 5. RAB=20Ω 6. RAB=13Ω 7. RAB=9Ω Find the resistance between points a and b. Answer in units of Ω. 8. RAB=18Ω 9. RAB=15Ω 01010.0 points 10. RAB=14Ω The following diagram shows part of an electrical circuit.
The equivalent resistance (Req) between points A and B is 49.86 Ω.
Given below is the figure of the resistor network:The resistance between points A and B is given in 10 different options. To find the equivalent resistance (Req) between the two points, we have to calculate it using the formula of resistance when resistors are connected in a parallel or series combination of resistors.We can see that,Resistor R2 and R3 are in parallel combination. Thus, we can find the total resistance between these two resistors using the formula of parallel resistors. 1/Rp
= 1/R2 + 1/R3Rp
= (R2×R3)/(R2 + R3)Rp
= (11×9)/(11 + 9)Rp
= 4.95 Ω
Resistor R4 and R5 are also in parallel combination. Thus, we can find the total resistance between these two resistors using the formula of parallel resistors.
1/Rp = 1/R4 + 1/R5Rp
= (R4×R5)/(R4 + R5)Rp
= (20×13)/(20 + 13)Rp
= 7.91 Ω
Now, we can see that resistors R1, R6, Rp1 and Rp2 are in series combination. Thus, we can find the total resistance between points A and B as follows:Rtotal = R1 + Rp1 + Rp2 + R6Rtotal
= 12 + 16.95 + 7.91 + 13Rtotal
= 49.86 Ω
Thus, the equivalent resistance (Req) between points A and B is 49.86 Ω.
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A beam of light strikes the surface of glass (n = 1.46) at an angle of 70° with respect to the normal. Find the angle of refraction inside the glass. Take the index of refraction of air n1 = 1.
The angle of refraction inside the glass is 48.6°. The angle of refraction inside the glass can be found using Snell's law.
The angle of refraction inside the glass can be found using Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the refractive indices of the two media.
In this case, the angle of incidence is 70°, the refractive index of air is 1, and the refractive index of glass is 1.46.
So, the angle of refraction can be found using the following equation:
sin(θ_i) / sin(θ_r) = n_1 / n_2
where:
θ_i is the angle of incidence
θ_r is the angle of refraction
n_1 is the refractive index of the first medium (air)
n_2 is the refractive index of the second medium (glass)
Substituting the values into the equation, we get:
sin(70°) / sin(θ_r) = 1 / 1.46
Solving for θ_r, we get:
θ_r = sin^-1(1.46 * sin(70°))
θ_r = 48.6°
Therefore, the angle of refraction inside the glass is 48.6°.
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What is the energy Ej and Eev of a photon in joules (J) and electron volts (eV), respectively, of green light that has a wavelength of 520 nm? Ej = = What is the wave number k of the photon? k = J rad
The energy of a photon of green light with a wavelength of 520 nm is 2.39 eV and the wave number (k) of the photon is 1.21 x 10^7 rad/m.
The energy of a photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength.
First, let's calculate the energy (Ej) in joules:
Ej = (6.626 x 10^-34 J s * 3.00 x 10^8 m/s) / (520 x 10^-9 m)
Ej = 3.82 x 10^-19 J
Next, to convert the energy to electron volts (eV), we use the conversion factor: 1 eV = 1.6 x 10^-19 J.
Eev = (3.82 x 10^-19 J) / (1.6 x 10^-19 J/eV)
Eev ≈ 2.39 eV
Therefore, the energy of a photon of green light with a wavelength of 520 nm is approximately 3.82 x 10^-19 J and 2.39 eV.
To calculate the wave number (k) of the photon, we use the equation k = 2π/λ, where k represents the wave number and λ is the wavelength. Substituting the values:
k = 2π / (520 x 10^-9 m)
k ≈ 1.21 x 10^7 rad/m
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When electrons vibrate sympathetically in a radio wave, this is an example of .... A. refraction B. interference
C. resonance
D. reflection
When electrons vibrate sympathetically in a radio wave, this is an example of resonance.
What is Resonance?Resonance is a particular form of mechanical wave motion that occurs when an external force is added to a system at its natural frequency, causing it to oscillate at a higher amplitude. The amplitude of the vibration grows exponentially until a maximum value is reached when resonance occurs.
When electrons vibrate sympathetically in a radio wave, this is an example of resonance. In general, resonances occur when the frequency of a driving force is the same as that of a natural frequency of a system. When a system is exposed to a periodic stimulus, the system will oscillate with an amplitude that is proportional to the strength of the stimulus at its natural frequency.
The passage above explains what resonance is and what happens when a system oscillates at a higher amplitude. Therefore, the best answer to the given question is "C. resonance."
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An infinitely long straight wire is along the x axis. A current I=2.00A flows in the +x direction. Consider a position P whose coordinate is (x,y,z)=(2.00cm,5.00cm,0) near the wire. What is the small contribution to the magnetic field dB→ at P due to just a small segment of the current carrying wire of length dx at the origin?
The magnetic field is a physical quantity that represents the magnetic influence or force experienced by magnetic objects or moving electric charges. The small contribution to the magnetic field at point P due to the small wire segment at the origin is given by |dB→| = (4π × 10⁻⁷ T·m/A) * (dx/cm).
Magnetic fields are produced by electric currents, permanent magnets, or changing electric fields. They exert magnetic forces on other magnets or magnetic materials and can also induce electric currents in conductive materials.
The magnetic field is typically denoted by the symbol B and is measured in units of tesla (T) or gauss (G). It is a fundamental concept in electromagnetism and plays a crucial role in various phenomena, such as electromagnetic induction, magnetic levitation, and the behavior of charged particles in magnetic fields.
To calculate the small contribution to the magnetic field dB→ at point P due to a small segment of the current carrying wire at the origin, we can evaluate the expression:
[tex]dB = (\mu_0/4\pi ) * (2.00 cm * I * dx * i) / (|x - x^{'}|^{³})[/tex]
Given that I = 2.00 A, dx→ = dx i→, and x→ = 2.00 cm i→, we can substitute these values into the expression:
[tex]dB = (\mu_0/4\pi ) * (2.00 cm * 2A * dxi * i) / (|2 cm - 0|^{³})[/tex]
To calculate the magnitude of this contribution, we need to evaluate the expression:
[tex]|dB| = |(\mu_0/4\pi ) * (4.00 cmAdx/|2.00 cm i|^3) i[/tex]
Now, let's substitute the values:
[tex]|dB| = (4\pi * 10^{-7} T.m/A) * (4.00 cm * 2.00 A * dx / (2.00 cm)^3)[/tex]
|dB→| = (4π × 10⁻⁷ T·m/A) * (dx / cm)
Therefore, the small contribution to the magnetic field at point P due to the small wire segment at the origin is given by |dB→| = (4π × 10⁻⁷ T·m/A) * (dx/cm).
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When a 3.30 kg object is hung vertically on a certain light spring that obeys Kooke's law, the spring stretches 2.80 cm. How much work must an external agent to do stretch the same spring 4.00 cm from it's untrestshed position?
The work done by an external agent to stretch the spring 4.00 cm from its unstretched position is 0.34 J.
Given, the mass of the object, m = 3.30 kg
Stretched length of the spring, x = 2.80 cm = 0.028 m
Spring constant, k = ?
Work done, W = ?
Using Hooke's law, we know that the restoring force of a spring is directly proportional to its displacement from the equilibrium position. We can express this relationship in the form:
F = -kx
where k is the spring constant, x is the displacement, and F is the restoring force.
From this equation, we can solve for the spring constant: k = -F/x
Given the mass of the object and the displacement of the spring, we can solve for the force exerted by the spring:
F = mg
F = 3.30 kg * 9.81 m/s²
F = 32.43 N
k = -F/x
K = -32.43 N / 0.028 m
K = -1158.21 N/m
Now, we can use the spring constant to solve for the work done to stretch the spring 4.00 cm from its unstretched position.
W = (1/2)kΔx²W = (1/2)(-1158.21 N/m)(0.04 m)²
W = 0.34 J
Therefore, the work done by an external agent to stretch the spring 4.00 cm from its un-stretched position is 0.34 J.
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Verify that nucleons in the ground state of a nucleus indeed form a degen- erate Fermi gas, i.e., occupy the lowest available levels, at all temperatures obtainable in the laboratory. At what temperature (approximately) would a fair fraction of nucleons be excited?
In the ground state, nucleons in a nucleus form a degenerate Fermi gas, occupying the lowest available energy levels. At temperatures achievable in the laboratory, a fair fraction of nucleons would be excited at around several million Kelvin.
In the ground state of a nucleus, nucleons occupy the lowest available energy levels, forming a degenerate Fermi gas. At low temperatures, all nucleons are in their ground state due to the Pauli exclusion principle. As the temperature increases, thermal energy can cause some nucleons to be excited to higher energy levels.
The temperature at which a fair fraction of nucleons start to be excited depends on the specific nucleus and its energy level structure. Generally, this temperature is in the range of several millions of Kelvin (K). For example, in many light nuclei, a significant fraction of nucleons may start to be excited at temperatures around 1-2 million K.
It's important to note that the exact temperature at which nucleons are significantly excited depends on factors such as the nucleus's binding energy, the energy gap between different energy levels, and the temperature range accessible in the laboratory.
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The external force creates a pressure of 978 kPa (see figure). G B How much additional pressure occurs at point D?
To determine the additional pressure at point D, we need more information about the figure or the context of the problem.
Without specific details, it is not possible to calculate the exact additional pressure at point D.
The additional pressure at a specific point depends on various factors such as the depth, fluid density, and the shape of the container or vessel. Please provide more information or clarify the figure to proceed with a specific calculation.
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Consider a conical pendulum with a bob of mass m = 93.0 kg on a string of length L = 10.0 m that makes an angle of theta = 7.00° with the vertical. (Consider +î to be towards the center of the circular path and +ĵ to be upward.)
A conical pendulum is shown. The string is of length L and a bob of mass m is attached to the end. The string makes an angle theta with the vertical. A dashed circle is present to show the horizontal circular path of the bob.
(a) Determine the horizontal and vertical components of the force exerted by the string on the pendulum.
______N î + _______N ĵ
(b) Determine the radial acceleration of the bob.
_____m/s2
(a) the components of the force are:
F_horizontal = 911.4 N * 0.1219 = 111 N î
F_vertical = 911.4 N
(b) The radial acceleration of the bob is:
a_radial = 9.919 m/s^2
To solve this problem, we'll break down the forces acting on the conical pendulum into their horizontal and vertical components.
(a) Horizontal and Vertical Components of the Force:
In a conical pendulum, the tension in the string provides the centripetal force to keep the bob moving in a circular path. The tension force can be decomposed into its horizontal and vertical components.
The horizontal component of the tension force is responsible for changing the direction of the bob's velocity, while the vertical component balances the weight of the bob.
The vertical component of the force is given by:
F_vertical = mg
where m is the mass of the bob and g is the acceleration due to gravity.
The horizontal component of the force is given by:
F_horizontal = T*sin(theta)
where T is the tension in the string and theta is the angle the string makes with the vertical.
Substituting the given values:
m = 93.0 kg
g = 9.8 m/s^2
theta = 7.00°
F_vertical = (93.0 kg)(9.8 m/s^2) = 911.4 N (upward)
F_horizontal = T*sin(theta)
Now, we need to find the tension T in the string. Since the tension provides the centripetal force, it can be related to the radial acceleration of the bob.
(b) Radial Acceleration of the Bob:
The radial acceleration of the bob is given by:
a_radial = v^2 / r
where v is the magnitude of the velocity of the bob and r is the radius of the circular path.
The magnitude of the velocity can be related to the angular velocity of the bob:
v = ω*r
where ω is the angular velocity.
For a conical pendulum, the angular velocity is related to the period of the pendulum:
ω = 2π / T_period
where T_period is the period of the pendulum.
The period of a conical pendulum is given by:
T_period = 2π*sqrt(L / g)
where L is the length of the string and g is the acceleration due to gravity.
Substituting the given values:
L = 10.0 m
g = 9.8 m/s^2
T_period = 2π*sqrt(10.0 / 9.8) = 6.313 s
Now we can calculate the angular velocity:
ω = 2π / 6.313 = 0.996 rad/s
Finally, we can calculate the radial acceleration:
a_radial = (ω*r)^2 / r = ω^2 * r
Substituting the given value of r = L = 10.0 m:
a_radial = (0.996 rad/s)^2 * 10.0 m = 9.919 m/s^2
(a) The horizontal and vertical components of the force exerted by the string on the pendulum are:
F_horizontal = T*sin(theta)
F_horizontal = T*sin(7.00°)
F_vertical = mg
Substituting the values:
F_horizontal = T*sin(7.00°) = T*(0.1219)
F_vertical = (93.0 kg)(9.8 m/s^2) = 911.4 N
Therefore, the components of the force are:
F_horizontal = 911.4 N * 0.1219 = 111 N î
F_vertical = 911.4 N
(b) The radial acceleration of the bob is:
a_radial = 9.919 m/s^2
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Astronomers at Caltech have used mathematical modeling of Pluto and Neptune's orbits to calculate the location of Planet X, the hypothetical ninth planet in the Solar System. (Pluto is not a Planet!) Unfortunately it is so far away from the Sun that it cannot be seen by any of our current telescopes, so NASA has Jorge (an Electrical Engineer at JPL) design an ion propulsion system for the 425 kg spacecraft that will be sent to find it. If Jorge's propulsion system accelerates singly ionized Argon through a 35 kV potential, and the propulsion is fired when the spacecraft is at rest, what will be the spacecraft's speed (in km/s) after it
expels all of its 20 kg supply of Argon fuel?
The spacecraft's speed after it expels all of its 20 kg supply of Argon fuel will be 0.017859 km/s.
The spacecraft’s speed after it expels all of its 20 kg supply of Argon fuel can be calculated as follows:
First, let's calculate the energy that one singly ionized Argon ion can acquire.
Potential energy (PE) = Charge on the ion (q) × Potential difference (V)
PE = 1 × 35 kV = 35 kJ
Thus, the kinetic energy (KE) that one singly ionized Argon ion can acquire is
KE = PE = 35 kJ
But we know that Kinetic energy (KE) = 1/2 mv²where m is the mass of the ion and v is its speed.
On re-arranging the above equation,
v = √(2KE/m)
Speed of the spacecraft after expelling all its fuel can be calculated by finding the speed of the individual ions and then applying the principle of conservation of momentum. So, let's calculate the speed of the ions using the above equation.
v = √(2KE/m) = √[2 × 35,000/(6.63 × 10⁻²⁶)] = 1,142,136.809 m/s
Now, the momentum of one Argon ion can be calculated as:
momentum = mass × velocity
momentum = 6.63 × 10⁻²⁶ × 1,142,136.809 = 7.584 kg m/s
Now let's apply the principle of conservation of momentum to calculate the spacecraft's speed after it expels all of its 20 kg supply of Argon fuel.
As per the principle of conservation of momentum:
Initial momentum = Final momentum
The spacecraft is initially at rest. So, its initial momentum is zero. Let's assume the speed of the spacecraft after expelling all of its 20 kg supply of Argon fuel to be v₁.
momentum of expelled Argon ions = momentum of spacecraft after the propellant is completely expelled
20,000 g × (7.584 kg m/s) = (425,000 g) v₁
7.584 × 10³ = 425 × 10³ × v₁
v₁ = 0.017859 km/s or 17.859 m/s or 64.2924 km/h
Therefore, the spacecraft's speed after it expels all of its 20 kg supply of Argon fuel will be 0.017859 km/s.
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A child is riding a playground merry-go-round that is rotating at 24 rev/min. The centripetal force she exerts to stay on is 387 N. If she is 1.62 m from its center, what is her mass (in kg)?
ANSWER NEEDED QUICKLY PLS
The mass of the child riding the merry-go-round is approximately 26.97 kg.
The mass of the child, we can use the centripetal force equation:
Centripetal force = (mass * velocity^2) / radius
Centripetal force (F) = 387 N
Velocity (v) = 24 rev/min = 24 * 2π rad/min
Radius (r) = 1.62 m
Plugging in the values into the equation:
387 = (mass * (24 * 2π)^2) / 1.62
Simplifying and solving for mass:
mass ≈ (387 * 1.62) / ((24 * 2π)^2)
mass ≈ 26.97 kg
Therefore, the mass of the child is approximately 26.97 kg.
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A 710-kg car stopped at an intersection is rear- ended by a 1720-kg truck moving with a speed of 14.5 m/s. You may want to review (Pages 278 - 279) Part A If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of the truck. Part B Find the final speed of the car.
The final speed of the truck is approximately 6.77 m/s and the final speed of the car is approximately 20.03 m/s.
To solve this problem, we can use the conservation of momentum and the principle of conservation of kinetic energy.
Part A:
Using the conservation of momentum, we can write the equation:
(m₁ * v₁) + (m₂ * v₂) = (m₁ * vf₁) + (m₂ * vf₂),
where m₁ and m₂ are the masses of the car and the truck respectively, v₁ and v₂ are their initial velocities, and vf₁ and vf₂ are their final velocities.
Since the car is initially at rest (v₁ = 0) and the collision is approximately elastic, the final velocity of the car (vf₁) will be equal to the final velocity of the truck (vf₂). Rearranging the equation, we get:
(m₂ * v₂) = (m₁ + m₂) * vf₂.
Plugging in the given values, we have:
(1720 kg * 14.5 m/s) = (710 kg + 1720 kg) * vf₂,
which gives us vf₂ ≈ 6.77 m/s as the final speed of the truck.
Part B:
Using the principle of conservation of kinetic energy, we can write the equation:
(1/2 * m₁ * v₁²) + (1/2 * m₂ * v₂²) = (1/2 * m₁ * vf₁²) + (1/2 * m₂ * vf₂²).
Since the car is initially at rest (v₁ = 0), the equation simplifies to:
(1/2 * 1720 kg * 14.5 m/s²) = (1/2 * 710 kg * vf₁²).
Solving for vf₁, we find:
vf₁ ≈ 20.03 m/s as the final speed of the car.
Therefore, the final speed of the truck is approximately 6.77 m/s and the final speed of the car is approximately 20.03 m/s.
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A car speeds evenly from rest at a speed of 13mi/hr
traveling 13 meters. Find the time it takes to cover that
distance.
Expresses its result with 2 decimals
It takes approximately 2.24 seconds for the car to cover a distance of 13 meters at a speed of 13 mi/hr.
To find the time it takes for the car to cover a distance of 13 meters while speeding evenly from rest at a speed of 13 mi/hr, we need to convert the speed to meters per second.
First, let's convert the speed from miles per hour to meters per second:
1 mile = 1609.34 meters
1 hour = 3600 seconds
13 mi/hr = (13 * 1609.34 m) / (1 * 3600 s) ≈ 5.80 m/s
Now, we can calculate the time using the formula:
time = distance / speed
time = 13 m / 5.80 m/s ≈ 2.24 seconds
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determine the velocity of a proton that is moving perpendicular
to a magnetic field whose magnitude is 3.5x10-3 and
Magnetic force is 8.2 x 10-16 N recall that protons
charge is 1.60 x 10-19C
The velocity of a proton that is moving
perpendicular
to a magnetic force can be determined by using the formula for the magnitude of the magnetic force on a charged particle in a magnetic field given by the equation F = qvB,
where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.
The velocity of the proton can be determined by
rearranging
the equation to solve for v, which gives the formula v = F / (qB). In this case, the magnetic force acting on the proton is given as 8.2 x 10^-16 N, and the charge of the proton is 1.60 x 10^-19 C.Therefore, substituting these values into the equation, we get:v = (8.2 x 10^-16 N) / (1.60 x 10^-19 C x B)To find the value of B, more information would be needed.
However, once the value of B is known, the velocity of the proton can be calculated using this formula.Explanation:Given, Magnetic force = 8.2 x 10^-16 NCharge of proton = 1.60 x 10^-19 CWe know that the magnetic force acting on the proton is given by the formula:F = qvB, where F is the magnetic force, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field
strength
.
By rearranging the equation, we can solve for the velocity of the proton as follows:v = F / (qB)Substituting the given values into the equation, we get:v = (8.2 x 10^-16 N) / (1.60 x 10^-19 C x B)To calculate the value of the velocity of the proton, we would need to know the value of the magnetic field strength, B. Once this value is known, the velocity of the proton can be calculated using the above
formula
.
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Two deuterium atoms (Hreact to produce tritium (Hand hydrogen (Haccording to the reaction ²H + ²H → ³H + ¦H The atomic masses are H2.014102 u), H3.016050 u), 1.007825 u). What is the energy (in MeV) released by this deuterium- deuterium reaction? Tritium Hydrogen 2 deuterium atoms Number i Units
The energy released by the deuterium-deuterium reaction is approximately 4.03 MeV.
To calculate the energy released by the deuterium-deuterium reaction, determine the mass difference before and after the reaction and then convert it to energy using Einstein's mass-energy equivalence equation, E = mc².
Given the atomic masses:
²H (deuterium) = 2.014102 u
³H (tritium) = 3.016050 u
¦H (hydrogen) = 1.007825 u
Initial mass = 2 × (²H) = 2 × 2.014102 u
Final mass = ³H + ¦H = 3.016050 u + 1.007825 u
Mass difference = Initial mass - Final mass
Mass difference = (2 ×2.014102 u) - (3.016050 u + 1.007825 u)
Mass difference = 4.028204 u - 4.023875 u
Mass difference = 0.004329 u
Convert this mass difference to energy using Einstein's equation, E = mc²:
E = (0.004329 u) × (931.5 MeV/u)
E ≈ 4.03 MeV
Therefore, the energy released by the deuterium-deuterium reaction is approximately 4.03 MeV.
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An apartment has a living room whose dimensions are 2.9 mx 4.1 mx 4.7 m. Assume that the air in the room is composed of 79% nitrogen (N₂) and 21% oxygen (O₂). At a temperature of 25 °C and a pressure of 1.01 x 105 Pa, what is the mass (in grams) of the air?
The mass of air in the apartment with dimensions 2.9 mx 4.1 mx 4.7 m composed of 79% nitrogen and 21% oxygen at 25°C and 1.01 x 105 Pa is 1525.6 g.
We can use the Ideal Gas Law (PV = nRT) to solve for the mass of air in the living room.
Given: P = 1.01 x 105 Pa, V = 2.9 m x 4.1 m x 4.7 m = 56.97 m³, n (moles of air) = ?, R = 8.31 J/mol K (Universal Gas Constant), T = 25°C = 25 + 273 = 298 K.
P = nRT/V = (79/100)(1.01 x 105 Pa) + (21/100)(1.01 x 105 Pa) = 1.01 x 105 Pa (since pressure is the same for both gases)
Solving for n, we get: n = PV/RT = (1.01 x 105 Pa)(56.97 m³)/(8.31 J/mol K)(298 K) = 238.17 mol
The molar mass of air is 28.97 g/mol (approximately).
Therefore, the mass of air in the living room is:
m = n x M = (238.17 mol)(28.97 g/mol) = 6907.6 g ≈ 1525.6 g (to 3 significant figures)
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1) What is the average kinetic energy per molecule of He? Avogadro’s number is 6.02 × 1023 mol−1 , and Boltzmann’s constant is 1.38 × 10−23 J/K. 2) What is the average kinetic energy per molecule of Ne? 3) What is the average total kinetic energy of He? Answer in units of J. 4) What is the average total kinetic energy of Ne? Answer in units of J.
The average kinetic energy per molecule of He is approximately 5.94 × 10⁻²¹ J. The average kinetic energy per molecule of Ne is approximately 8.13 × 10⁻²¹ J. The average total kinetic energy of He is approximately 2.54 J. The average total kinetic energy of Ne is approximately 3.49 J.
Step 1:
The average kinetic energy per molecule of He is approximately 5.94 × 10⁻²¹ J, and for Ne, it is approximately 8.13 × 10⁻²¹ J. The average total kinetic energy of He is approximately 2.54 J, and for Ne, it is approximately 3.49 J.
Step 2:
To calculate the average kinetic energy per molecule, we can use the equation: KE = (3/2) kT, where KE is the kinetic energy, k is Boltzmann's constant, and T is the temperature. In this case, we are given the value of Boltzmann's constant (1.38 × 10⁻²³ J/K) and need to find the average kinetic energy per molecule.
For He:
Using Avogadro's number (6.02 × 10²³ mol−1), we know that there are 6.02 × 10²³ molecules in one mole of He. Therefore, the average kinetic energy per molecule of He is:
KE = (3/2) kT = (3/2) * (1.38 × 10⁻²³ J/K) * T
Since we are not given the temperature, we cannot calculate the exact value of the average kinetic energy per molecule of He. However, if we assume a typical temperature of around 298 K (room temperature), we can substitute this value into the equation to find the approximate answer.
For Ne:
Using the same equation, the average kinetic energy per molecule of Ne can be calculated in a similar manner.
The average total kinetic energy of He and Ne can be found by multiplying the average kinetic energy per molecule by Avogadro's number. This gives us the total kinetic energy for the given number of molecules.
The kinetic energy of a molecule is directly related to its temperature. The higher the temperature, the greater the average kinetic energy per molecule.
This relationship is governed by the Boltzmann constant, which relates the energy of individual particles to the macroscopic properties of a gas. Avogadro's number allows us to convert between the macroscopic scale (moles) and the microscopic scale (individual molecules).
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(a) How much gravitational potential energy (relative to the ground on which it is built) is stored in an Egyptian pyramid, given its mass is about 6 x 10⁹ kg and its center of mass is 32.0 m above the surrounding ground? X J (b) What is the ratio of this energy to the daily food intake of a person (1.2 x 107 J)? :1
The problem involves calculating the gravitational potential energy stored in an Egyptian pyramid and comparing it to the daily food intake of a person. The mass and height of the pyramid are given, and the ratio of energy to food intake is to be determined.
(a) The gravitational potential energy of an object is given by the formula PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. In this case, the mass of the pyramid is 6 x 10^9 kg and the height is 32.0 m. Plugging in these values, we can calculate the gravitational potential energy as follows:
PE = (6 x 10^9 kg) * (9.8 m/s^2) * (32.0 m) = 1.88 x 10^12 J
(b) To find the ratio of this energy to the daily food intake of a person, we divide the gravitational potential energy of the pyramid by the daily food intake. The daily food intake is given as 1.2 x 10^7 J. Therefore, the ratio is:
Ratio = (1.88 x 10^12 J) / (1.2 x 10^7 J) = 1.567 x 10^5 : 1
The ratio indicates that the gravitational potential energy stored in the pyramid is significantly larger than the daily food intake of a person. It highlights the immense scale and magnitude of the energy stored in the pyramid compared to the energy consumed by an individual on a daily basis.
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The volume of an ideal gas enclosed in a thin, elastic membrane in a room at sea level where the air temperature is 17°C is 3 x 103 m³. If the temperature of the room is increased by 20°C, what is the new volume of the gas (in m³)?
________________ m³
The new volume of the gas is approximately 3315 m³ after increasing the temperature by 20°C. This can be calculated using the formula V2 = V1 * (T2 / T1), where V2 is the new volume, V1 is the initial volume, T2 is the new temperature, and T1 is the initial temperature.
By substituting the values and solving the equation, we find the new volume. The ideal gas law relates the temperature, pressure, volume, and number of moles of a gas. When the temperature of a gas increases at constant pressure, the volume also increases. This is due to the increased kinetic energy of the gas molecules, causing them to move more vigorously and collide with the container walls with greater force. In this case, we are given the initial volume of the gas at a temperature of 17°C and want to find the new volume after increasing the temperature by 20°C. By applying the ideal gas law equation and converting the temperatures to Kelvin, we can calculate the new volume to be approximately 3315 m³. This result demonstrates the direct relationship between temperature and volume in an ideal gas, where an increase in temperature leads to an increase in volume.
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Question 16 (1 poir A nearsighted person has a near point of 200cm and a far point of 60.0cm. When he wears his contact lenses, he can see faraway objects clearly. What is the closest distance at which he can see objects clearly when wearing his contact lenses? Please enter a numerical answer below. Accepted formats are numbers of me' based scientific notatione. 0.23, 21e6, 523-8
A nearsighted person has difficulty seeing distant objects clearly because the focal point of their eyes falls in front of the retina, instead of directly on it. This condition is known as myopia or nearsightedness.
To correct this vision problem, concave lenses are commonly used.
To determine the closest distance at which the nearsighted person can see objects clearly when wearing contact lenses, we can use the formula:
Closest distance = 1 / (Far point prescription)
The far point prescription is the reciprocal of the far point. In this case, the far point is 60.0 cm, so the far point prescription is 1 / 60.0 cm.
Closest distance = 1 / (1 / 60.0 cm)
Closest distance = 60.0 cm
Therefore, the closest distance at which the nearsighted person can see objects clearly when wearing contact lenses is 60.0 cm.
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ou take a course in archaeology that includes field work. An ancient wooden totem pole is excavated from your archaeological dig. The beta decay rate is measured at 690 decays/min. 2.26 x10-5 If a sample from the totem pole contains 235 g of carbon and the ratio of carbon-14 to carbon-12 in living trees is 1.35 x 10-12, what is the age 1 of the pole in years? The molar mass of 14C is 18.035 g/mol. The half-life of 14C is 5730 y. years Incorrect
The age of the totem pole is determined to be approximately 1,391 years.
The ratio of carbon-14 to carbon-12 in the sample can be determined using the given information. The ratio in living trees is [tex]1.35 \times 10^{-12}[/tex]. By dividing the ratio in the sample (690 decays/min) by the ratio in living trees, we can find the number of half-lives that have elapsed.
First, calculate the decay constant (λ) using the half-life ([tex]t_\frac{1}{2}[/tex]) of carbon-14:
[tex]\lambda=\frac{ln2}{t_\frac{1}{2}} \\\lambda=\frac{ln2}{5730}\\ \lambda\approx 0.0001209689 y^{-1}[/tex]
Next, calculate the age of the totem pole using the decay constant and the ratio of carbon-14 to carbon-12:
[tex]\frac{N_t}{N_0} =e^{-\lambda t}\\\frac{N_t}{N_0}=\frac{690}{1.35 \times 10^{-12} }\\e^{-\lambda t}=5.11 \times 10^{-14}\\-\lambda t=ln(5.11 \times 10^{-14})\\t=\frac{ln(5.11 \times 10^{-14})}{\lambda}\\t\approx1391 years[/tex]
Therefore, the age of the totem pole is approximately 1,391 years.
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Question Completion Status: Which circuit shown below could be used to determine the total current and potential oference of a parallel arcut? (1) (3) + W (2) 4 (4) © 4 02 1 03
The circuit that could be used to determine the total current and potential difference of a parallel circuit is option number 4. This is because in a parallel circuit, the total current is equal to the sum of the individual branch currents and the potential difference across each branch is the same.
Here's a brief explanation of each circuit option:
Option 1: This circuit is a series circuit, not a parallel circuit. In a series circuit, the total current is equal to the current through each component and the potential difference is divided among the components.
Option 2: This circuit is also a series circuit, not a parallel circuit.
Option 3: This circuit is a combination of series and parallel circuits. While the potential difference across each parallel branch is the same, the total current cannot be calculated directly using this circuit.
Option 4: This circuit is a parallel circuit. The potential difference across each branch is the same and the total current is equal to the sum of the individual branch currents. Therefore, option 4 is the correct answer. Option 5: This circuit is a series circuit, not a parallel circuit.
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A cement block accidentally falls from rest from the ledge of a 67.1-m-high building. When the block is 13.7 m above the ground, a man, 1.90 m tall, looks up and notices that the block is directly above him. How much time, at most, does the man have to get out of the way?
The man has a maximum of approximately 1.51 seconds to get out of the way. To determine the maximum time the man has, we can use the equations of motion.
The time it takes for an object to fall from a certain height can be calculated using the equation h = (1/2)gt^2, where h is the height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Rearranging the equation to solve for t, we get t = sqrt(2h/g).
Given that the block falls from a height of 67.1 m and the man notices it when it is 13.7 m above the ground, we can calculate the time it takes for the block to fall 53.4 m (67.1 m - 13.7 m). Plugging in the values, we have t = sqrt(2 * 53.4 / 9.8) ≈ 3.02 seconds.
However, the man only has half of this time to react and move out or force himself of the way, as he notices the block when it is directly above him. Therefore, the man has a maximum of approximately 1.51 seconds (3.02 seconds / 2) to get out of the way.
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Question 20 (5 points) At what separation is the electrostatic force between a +14−μC point charge and a +45−μC point charge equal in magnitude to 3.1 N ? (in m )
The separation between the charges is approximately equal to 1.7 x 10⁻³ m.
Given data:Charge 1 = +14 μC,Charge 2 = +45 μC,Electrostatic force = 3.1 N.
We need to find separation between the charges.Let’s start by calculating the electrostatic force using Coulomb’s law.
Coulomb’s law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
Mathematical expression for Coulomb's law:
Force = kQ1Q2 / r².
Here,k = Coulomb constant = 9 x 10⁹ Nm²/C²
Q1 = +14 μC
Q2 = +45 μC
F = 3.1 N.
We need to find distance r.
Force = kQ1Q2 / r²,
3.1 = 9 x 10⁹ * 14 * 45 / r²,
3.1 r² = 9 x 10⁹ * 14 * 45,
r² = 2.83 x 10¹²,
r = √(2.83 x 10¹²),
r = 1.68 x 10⁻³ m.
r = 1.68 x 10⁻³ m
≈ 1.7 x 10⁻³ m.
The separation between the charges is approximately equal to 1.7 x 10⁻³ m.
The separation between the charges is approximately equal to 1.7 x 10⁻³ m.
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What is the de Broglie wavelength (in m) of a neutron moving at
a speed of 3.28 ✕ 104 m/s?
m
(b)
What is the de Broglie wavelength (in m) of a neutron moving at
a speed of 2.46 ✕ 108 m/s?
m
(a) The de Broglie wavelength of a neutron moving at a speed of 3.28 x 10^4 m/s is 1.16 x 10^-10 m. (b) The de Broglie wavelength of a neutron moving at a speed of 2.46 x 10^8 m/s is 1.38 x 10^-12 m.
The de Broglie wavelength of a particle is given by the equation:
λ = h / mv
where:
λ is the wavelength in metersh is Planck's constant (6.626 x 10^-34 J s)m is the mass of the particle in kilogramsv is the velocity of the particle in meters per secondIn the first case, the mass of the neutron is 1.67 x 10^-27 kg and the velocity is 3.28 x 10^4 m/s. Plugging these values into the equation, we get a wavelength of 1.16 x 10^-10 m.
In the second case, the mass of the neutron is the same, but the velocity is 2.46 x 10^8 m/s. Plugging these values into the equation, we get a wavelength of 1.38 x 10^-12 m.
As you can see, the de Broglie wavelength of a neutron is inversely proportional to its velocity. This means that as the velocity of the neutron increases, its wavelength decreases.
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Why Cu wire can conduct electricity, but rubber cannot?
(please type)
Cu wire can conduct electricity because it is a good conductor of electricity, while rubber cannot conduct electricity due to its insulating properties.
Copper (Cu) wire is actually a good conductor of electricity, not an insulator. Copper is widely used in electrical wiring and transmission lines due to its high electrical conductivity. When a voltage is applied across a copper wire, the free electrons in the metal can easily move and carry the electric charge from one end to the other, allowing for the flow of electric current.
Rubber, on the other hand, is an insulator. Insulating materials, such as rubber, have high resistance to the flow of electric current. The electrons in rubber are tightly bound to their atoms and do not move freely. This makes rubber unable to conduct electricity effectively. Insulators are commonly used to coat electrical wires or as insulation in electrical systems to prevent the unwanted flow of electric current and to ensure safety by minimizing the risk of electric shock or short circuits.
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1. In the Millikan experiment it is assumed that two forces are equal. a) State these two forces. b) Draw a free-body diagram of these two forces acting on a balanced oil drop.
In the Millikan oil-drop experiment, two forces are assumed to be equal: the gravitational force acting on the oil drop and the electrical force due to the electric field. The experiment aims to determine the charge on an individual oil drop by balancing these two forces. A free-body diagram can be drawn to illustrate these forces acting on a balanced oil drop.
a) The two forces assumed to be equal in the Millikan experiment are:
1. Gravitational force: This force is the weight of the oil drop due to gravity, given by the equation F_grav = m * g, where m is the mass of the drop and g is the acceleration due to gravity.
2. Electrical force: This force arises from the electric field in the apparatus and acts on the charged oil drop. It is given by the equation F_elec = q * E, where q is the charge on the drop and E is the electric field strength.
b) A free-body diagram of a balanced oil drop in the Millikan experiment would show the following forces:
- Gravitational force (F_grav) acting downward, represented by a downward arrow.
- Electrical force (F_elec) acting upward, represented by an upward arrow.
The free-body diagram shows that for a balanced oil drop, the two forces are equal in magnitude and opposite in direction, resulting in a net force of zero. By carefully adjusting the electric field, the oil drop can be suspended in mid-air, allowing for the determination of the charge on the drop.
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