The magnitude of the deflection on the screen caused by the Earth's gravitational field can be calculated as below: F_gravity = m * g, where m = mass of electron, g = acceleration due to gravity = 9.8 m/s².
F_gravity = 9.1 x 10⁻³¹ kg * 9.8 m/s² = 8.91 x 10⁻³⁰ N Force on the electron will be F = q * E, where q = charge on electron = 1.6 x 10⁻¹⁹ C, E = electric field = V / d, where V = accelerating voltage = 2.45 x 10³ V, d = distance from the electron gun to the screen = 36.6 cm = 0.366 m.
E = V / d = 2.45 x 10³ V / 0.366 m = 6.68 x 10³ V/mF = q * E = 1.6 x 10⁻¹⁹ C * 6.68 x 10³ V/m = 1.07 x 10⁻¹⁵ N Force on the electron due to the Earth's gravitational field = F_gravity = 8.91 x 10⁻³⁰ NNet force on the electron = F_net = √(F_gravity² + F²)F_net = √(8.91 x 10⁻³⁰ N)² + (1.07 x 10⁻¹⁵ N)² = 1.07 x 10⁻¹⁵ NAngle of deflection = tan⁻¹(F_gravity / F) = tan⁻¹(8.91 x 10⁻³⁰ / 1.07 x 10⁻¹⁵) = 0.465°Magnitude of deflection = F_net * d / (q * V) = 1.07 x 10⁻¹⁵ N * 0.366 m / (1.6 x 10⁻¹⁹ C * 2.45 x 10³ V) = 1.47 x 10⁻³ mm(b) The direction of the deflection on the screen caused by the Earth's gravitational field is down.
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Consider the RLC circuit shown in the figure. w R V Select 'True, "False' or 'Cannot tell' for the following statements. The current through the inductor is the same as the current through the resistor at all times. The current through the inductor always equals the current charging/discharging the capacitor. The voltage drop across the resistor is the same as the voltage drop across the inductor at all times. Energy is dissipated in the resistor but not in either the capacitor or the inductor. Submit Answer Tries 0/12 What is the value of the inductance L so that the above circuit carries the largest current? Data: R = 2.39x102 12, f = 1.65x103 Hz, C = 6.10x10-3 F, Vrms = 9.69x101 v. Submit Answer Tries 0/12 Using the inductance found in the previous problem, what is the impedance seen by the voltage source? Submit Answer Tries 0/12
Statement 1: False. The current through the inductor is not always the same as the current through the resistor. It depends on the frequency and phase difference between the voltage source and the circuit components.
Statement 2: Cannot tell. The current through the inductor can be different from the current charging/discharging the capacitor depending on the frequency and phase relationship between the components.
Statement 3: False. The voltage drop across the resistor is not always the same as the voltage drop across the inductor. It depends on the frequency and phase relationship between the components.
Statement 4: False. Energy is dissipated in the resistor, but energy can also be stored and released in the capacitor and inductor as they store electrical energy in their electric and magnetic fields, respectively.
Regarding the value of inductance L that carries the largest current, the information provided (R, f, C, Vrms) is not sufficient to determine it.
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5. [-/2 Pointsj DETAILS SERFICI0 10.3.0 Vehn A disk 7.90 cm in radius rotates at a constant rate of 1 140 rev/min about its central axis. (a) Determine its angular speed. rad/s (b) Determine the tangential speed at a point 3.08 cm from its center. m/s (c) Determine the radial acceleration of a point on the rim. magnitude km/s2 direction -Select- (d) Determine the total distance a point on the rim moves in 1.92 s. m Need Help? Read It Master it
The question involves a disk with a radius of 7.90 cm rotating at a constant rate of 1,140 rev/min about its central axis. The task is to determine the angular speed, tangential speed at a specific point, radial acceleration at the rim, and the total distance traveled by a point on the rim in a given time.
(a) To find the angular speed, we need to convert the given rate from revolutions per minute (rev/min) to radians per second (rad/s). Since one revolution is equivalent to 2π radians, we can calculate the angular speed using the formula: angular speed = (2π * rev/min) / 60. Substituting the given value of 1,140 rev/min into the formula will yield the angular speed in rad/s.
(b) The tangential speed at a point on the disk can be calculated using the formula: tangential speed = radius * angular speed. Given that the radius is 3.08 cm, and we determined the angular speed in part (a), we can substitute these values into the formula to find the tangential speed in m/s.
(c) The radial acceleration of a point on the rim can be determined using the formula: radial acceleration = (tangential speed)^2 / radius. Substituting the tangential speed calculated in part (b) and the given radius, we can calculate the magnitude of the radial acceleration. However, the question does not provide the direction of the radial acceleration, so it remains unspecified.
(d) To determine the total distance a point on the rim moves in 1.92 s, we can use the formula: distance = tangential speed * time. Since we know the tangential speed from part (b) and the given time is 1.92 s, we can calculate the total distance traveled by the point on the rim.
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12. (II) (a) Show that the nucleus Be (mass = 8.005308 u) is unstable to decay into two a particles. (b) Is 'C stable against decay into three a particles? Show why or why not. tum what off b SECTIONS
(a) To determine the stability of the Be nucleus against decay into two alpha particles, we must compute the mass of the products (2 alpha particles) and compare it to the mass of the Be nucleus. Two alpha particles are equivalent to a helium nucleus. The mass of the helium nucleus is 4.001506 u. Therefore, the mass of two alpha particles is 8.003012 u.
The difference between the mass of the Be nucleus and the mass of two alpha particles is:Δm = M(Be) - M(2α) = 8.005308 u - 8.003012 u= 0.002296 u The decay into two alpha particles can proceed if the Q-value of the reaction is positive. The Q-value of the reaction is: Q = Δm c² = 0.002296 u x (1.6606 x 10-27 kg/u) x (2.998 x 108 m/s)²Q = 4.13 x 10-12 J This is a small amount of energy.
Therefore, the Be nucleus is unstable against decay into two alpha particles.(b) The carbon-12 nucleus is stable against decay into three alpha particles. To show why, we must compute the Q-value of the reaction. Three alpha particles are equivalent to a helium nucleus. The mass of the helium nucleus is 4.001506 u.
Therefore, the mass of three alpha particles is 12.004518 u. The difference between the mass of the C nucleus and the mass of three alpha particles is: Δm = M(C) - M(3α) = 12.000 u - 12.004518 u= -0.004518 u The decay into three alpha particles can proceed if the Q-value of the reaction is positive. The Q-value of the reaction is:
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A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. F…(N) (a) Find the work done by the force on the object as it moves from x=0 to x=5.00 m. ] (b) Find the work done by the force on the object as it moves from x=5.00 m to x=10.0 m. J (c) Find the work done by the force on the object as it moves from x=10.0 m to x=17.0 m. ] (d) If the object has a speed of 0.550 m/s at x=0, find its speed at x=5.00 m and its speed at x=17.0 m. speed at x=5.00 m m/s speed at x=17.0 m m/s
The work done by the force on the object as it moves from x=10.0 m to x=17.0 m is -267 J.
a) The work done by the force on the object as it moves from x=0 to x=5.00 m.The work done by the force on the object is equal to the change in the object's kinetic energy. In this case, the object's initial speed is zero. Hence, the work done by the force is equal to the kinetic energy that the object will have after moving a distance of 5.00 m.
Work done = ΔKE= (1/2)mv
2 - 0 = (1/2)(3.00 kg)(7.0 m/s)2
= 73.5 J
b) The work done by the force on the object as it moves from x=5.00 m to x=10.0 m.The work done by the force on the object is equal to the change in the object's kinetic energy. In this case, the object's initial speed is 7.0 m/s. Hence, the work done by the force is equal to the kinetic energy that the object will have after moving a distance of 5.00 m.
Work done = ΔKE
= (1/2)mv
2f - (1/2)mv2i= (1/2)(3.00 kg)(12.0 m/s)2 - (1/2)(3.00 kg)(7.0 m/s)2
= 210 J
c) The work done by the force on the object as it moves from x=10.0 m to x=17.0 m.The work done by the force on the object is equal to the change in the object's kinetic energy. In this case, the object's initial speed is 12.0 m/s. Hence, the work done by the force is equal to the kinetic energy that the object will have after moving a distance of 7.00 m.
Work done = ΔKE= (1/2)mv
2f - (1/2)mv2i= (1/2)(3.00 kg)(6.70 m/s)2 - (1/2)(3.00 kg)(12.0 m/s)2= -267 J (negative work as the force and displacement are in opposite directions)
Thus, the work done by the force on the object as it moves from x=0 to x=5.00 m is 73.5 J, the work done by the force on the object as it moves from x=5.00 m to x=10.0 m is 210 J and the work done by the force on the object as it moves from x=10.0 m to x=17.0 m is -267 J.
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A-Calculate the change in air pressure you will experience if you climb a 1000 m mountain, assuming for simplicity that the temperature and air density do not change over this distance and that they were 22 ∘C and 1.2 kg/m3 respectively, at the bottom of the mountain. Express your answer with the appropriate units. Enter negative value if the pressure decreases and positive value if the pressure increases.
b-
If you took a 0.45 LL breath at the foot of the mountain and managed to hold it until you reached the top, what would be the volume of this breath when you exhaled it there?
Express your answer with the appropriate units.
a) The change in air pressure when climbing a 1000 m mountain is -11,760 Pa (pressure decreases).
b) The volume of the breath when exhaled at the top of the mountain depends on the initial pressure and the pressure at the top, which requires further calculation based on the given values.
a) To calculate the change in air pressure as you climb a 1000 m mountain, we can use the hydrostatic pressure equation:
ΔP = -ρgh
where ΔP is the change in pressure, ρ is the air density, g is the acceleration due to gravity, and h is the change in height.
Given:
ρ = 1.2 kg/m^3
g = 9.8 m/s^2
h = 1000 m
Substituting these values into the equation, we get:
ΔP = -(1.2 kg/m^3)(9.8 m/s^2)(1000 m) = -11,760 Pa
Therefore, the change in air pressure is -11,760 Pa, indicating a decrease in pressure as you climb the mountain.
b) To calculate the volume of the breath when you exhale it at the top of the mountain, we can use Boyle's law, which states that the volume of a gas is inversely proportional to its pressure when temperature is constant:
P1V1 = P2V2
Given:
P1 = Initial pressure (at the foot of the mountain)
V1 = Initial volume (0.45 L)
P2 = Final pressure (at the top of the mountain)
V2 = Final volume (to be determined)
We can rearrange the equation to solve for V2:
V2 = (P1V1) / P2
The pressure at the top of the mountain can be calculated using the ideal gas law:
P2 = (ρRT) / M
where ρ is the air density, R is the ideal gas constant, T is the temperature, and M is the molar mass of air.
Given:
ρ = 1.2 kg/m^3
R = 8.314 J/(mol·K)
T = 22°C = 295 K (converted to Kelvin)
M = molar mass of air ≈ 28.97 g/mol
Substituting these values into the equation, we can calculate P2:
P2 = (1.2 kg/m^3)(8.314 J/(mol·K))(295 K) / (28.97 g/mol) ≈ 1205 Pa
Now we can substitute the values of P1, V1, and P2 into the equation for V2:
V2 = (P1V1) / P2 = (P1)(0.45 L) / P2
Substituting the appropriate values, we can calculate V2.
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The prescriber orders 30 mg of phenytoin (Dilantin) q8h po to be given to a toddler. The child weighs 44 pounds. The Pediatric reference book states that the maximum safe dose is 5 mg/kg/day. You have a bottle that has liquid phenytoin labeled 6 mg/mL. What is the calculated dosage you will give?
The correct calculated dose of phenytoin to be given to the pediatric patient is 5 mL.
Prescribed dose: 30 mg q8h po
Child's weight: 44 pounds
Maximum safe dose: 5 mg/kg/day
Liquid phenytoin concentration: 6 mg/mL
Convert the weight from pounds to kilograms.
Weight in kilograms = 44 pounds / 2.2 = 20 kg
Determine the maximum permissible dosage for the child.
Maximum safe dose = 5 mg/kg/day x 20 kg = 100 mg/day
Check if the prescribed dose is safe:
Prescribed dose per day = 30 mg x 3 = 90 mg/day
Since the prescribed dose (90 mg/day) is less than the maximum safe dose (100 mg/day), it is safe.
Calculate the calculated dose to be given:
Calculated dose = (Prescribed dose / Concentration) x Dosage form
Prescribed dose = 30 mg
Concentration = 6 mg/mL
Dosage form = mL
Calculating the calculated dose:
Calculated dose = (30 mg / 6 mg/mL) x mL
Calculated dose = 5 mL
Therefore, the correct calculated dose of phenytoin to be given to the pediatric patient is 5 mL.
Please note that this answer assumes the prescribed dose of 30 mg q8h po means 30 mg every 8 hours orally. If there are any specific instructions or considerations from the healthcare provider, they should be followed.
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Question 15 1 pts A spherical drop of water in air acts as a converging lens. How about a spherical bubble of air in water? It will Act as a converging lens Not act as a lens at all Act as a diverging
The correct option is "Act as a diverging".
Detail Answer:When a spherical bubble of air is formed in water, it behaves as a diverging lens. As it is a lens made of a convex shape, it diverges the light rays that come into contact with it. Therefore, a spherical bubble of air in water will act as a diverging lens.Lens is a transparent device that is used to refract or bend light.
There are two types of lenses, i.e., convex and concave. Lenses are made from optical glasses and are of different types depending upon their applications.Lens works on the principle of refraction, and it refracts the light when the light rays pass through it. The lenses have an axis and two opposite ends.
The lens's curved surface is known as the radius of curvature, and the center of the lens is known as the optical center . The type of lens depends upon the curvature of the surface of the lens. The lens's curvature surface can be either spherical or parabolic, depending upon the type of lens.
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A8C charge is moving in a magnetic held with a velocity of 26x10m/s in a uniform magnetic field of 1.7. the velocity vector is making a 30° angle win the direction of magnetic field, what is the magnitude of the force experienced by the charge
The magnitude of the force experienced by the charge in a magnetic field with a velocity of 26 x 10 m/s is 932.8 N
We are given the following information in the question:
Charge on the moving charge, q = 8 C
The velocity of the charge, v = 26 × 10 m/s
Magnetic field strength, B = 1.7 T
The angle between the velocity vector and magnetic field direction, θ = 30°
We can use the formula for the magnitude of the magnetic force experienced by a moving charge in a magnetic field, which is : F = qvb sin θ
where,
F = force experienced by the charge
q = charge on the charge
m = mass of the charge
n = number of electrons
v = velocity of the charger
b = magnetic field strength
θ = angle between the velocity vector and magnetic field direction
Substituting the given values, we get :
F = (8 C)(26 × 10 m/s)(1.7 T) sin 30°
F = (8)(26 × 10)(1.7)(1/2)F = 932.8 N
Thus, the magnitude of the force experienced by the charge is 932.8 N.
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Q20) A block of mass [m] kg, moving on a horizontal frictionless surface with a speed [v] m/s, makes a collection with a block of mass [M] kg at rest. After the collision, the [m] kg block recoils with speed V=1.2m/s to the left. Find the speed V (in meters) of the [M] kg after collision. m m = 3.60 kg Vi = 4.60 m/s M = 8.45 kg Vf = 1.2 m/s M = before = ve m M after
Given the information provided:
Mass of block 1 (m1) = 3.6 kg
Speed of block 1 before collision (u) = 4.6 m/s
Speed of block 1 after collision (v1) = -1.2 m/s
Mass of block 2 (m2) = 8.45 kg
Speed of block 2 after collision (v2) = ?
Using the principle of conservation of momentum, we can set up the equation:
m1u1 + m2u2 = m1v1 + m2v2
Substituting the given values:
(3.6)(4.6) = (3.6)(-1.2) + (8.45)(v2) + 0
Simplifying:
16.56 = -4.32 + 8.45v2
Solving for v2:
8.45v2 = 16.56 + 4.32
8.45v2 = 20.88
v2 = 20.88 / 8.45
v2 = 2.47 m/s
Therefore, the speed of block 2 after the collision is 2.47 m/s.
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Two speakers each produce the same tone, in phase with each other. One is positioned at the origin, and one located on the x axis at the position x = 5 m. You are standing on the y axis at y = 7 m. If you walk a little bit in any direction, the sound intensity decreases. Are you at a position of constructive, destructive, or interference? Find 3 different possibilities for frequency of the tone emitted by the speaker Note: the speed of sound in air is 343 m/s
Summary: Based on the given information, the listener is at a position of destructive interference. Three different possibilities for the frequency of the tone could be determined using the formula for destructive interference .
Explanation: Destructive interference occurs when two waves with the same frequency and opposite phases meet and cancel each other out. In this scenario, the listener is positioned on the y-axis at y = 7 m, while the two speakers are located at the origin (0, 0) and on the x-axis at x = 5 m. Since the speakers are in phase with each other, the listener experiences the phenomenon of destructive interference.
To find the frequencies that result in destructive interference at the listener's position, we can use the formula for the path length difference (ΔL) between the two speakers:
ΔL = sqrt((x₁ - x)² + y²) - sqrt(x² + y²)
where x₁ represents the position of the second speaker (x₁ = 5 m) and x represents the listener's position on the x-axis.
Since the sound intensity decreases when the listener walks away from the origin, the path length difference ΔL should be equal to an odd multiple of half the wavelength (λ/2) to achieve destructive interference. The relationship between wavelength, frequency, and the speed of sound is given by the equation v = fλ, where v is the speed of sound in air (343 m/s).
By rearranging the formula, we have ΔL = (2n + 1)(λ/2), where n is an integer representing the number of half wavelengths.
Substituting the values into the equation, we can solve for the frequency (f):
ΔL = sqrt((5 - x)² + 7²) - sqrt(x² + 7²) = (2n + 1)(λ/2) 343/f = (2n + 1)(λ/2)
By considering three different values of n (e.g., -1, 0, 1), we can calculate the corresponding frequencies using the given formula and the speed of sound.
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A 0.05 kg steel ball and a 0.15-kg iron ball are moving in opposite directions and are on a head-on collision course. They both have a speed of 2.5 m/s and the collision will be elastic Calculate the final velocities of the balls and describe their motion
The final velocity of steel ball is 2.5m/s and of iron ball is -0.833m/s. The steel ball moves with a uniform motion while the iron ball moves with accelerated motion.
Given data:
Mass of steel ball, m1 = 0.05 kg
Mass of iron ball, m2 = 0.15 kg
Velocity of steel ball, u1 = 2.5 m/s
Velocity of iron ball, u2 = -2.5 m/s (opposite direction)
Collision type, elastic
Here, as both the balls are moving in the opposite direction, so the relative velocity will be equal to the sum of velocities of both the balls.
On collision, the balls will move away from each other with some final velocities (v1, v2) which can be calculated using the law of conservation of momentum, which states that,
Total initial momentum of the system = Total final momentum of the systemInitial momentum
= m1u1 + m2u2
= 0.05 × 2.5 + 0.15 × (-2.5)
= -0.125 kg m/s (negative sign indicates that momentum is in the opposite direction)
Final momentum = m1v1 + m2v2
Let's substitute the given values in the above equation.
-0.125 = 0.05 v1 + 0.15 v2 ... Equation (1)
Now, using the law of conservation of energy, which states that,
Total initial energy of the system = Total final energy of the system
Total initial kinetic energy of the system can be calculated as,
K.E. = 1/2 m1 u1² + 1/2 m2 u2²
= 1/2 × 0.05 × (2.5)² + 1/2 × 0.15 × (-2.5)²
= 0.625 J
Total final kinetic energy of the system can be calculated as,
K.E. = 1/2 m1 v1² + 1/2 m2 v2²
Now, let's substitute the given values in the above equation.
0.625 = 1/2 × 0.05 v1² + 1/2 × 0.15 v2² ... Equation (2)
Solving equation (1) and equation (2), we get:
v1 = 2.5 m/s (final velocity of steel ball)
v2 = -0.833 m/s (final velocity of iron ball)
As we can see that after collision, the steel ball moves away in the same direction as it was moving initially while the iron ball moves away in the opposite direction.
Hence, we can say that the steel ball moves with a uniform motion (constant velocity) while the iron ball moves with accelerated motion (decreasing velocity). The steel ball will continue to move with a velocity of 2.5 m/s in the forward direction and the iron ball will slow down to a stop and reverse its direction of motion.
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QUESTION 6 Determine the diode voltage and current using a piecewise linear model if the diode parameters are Vp = 0.8 V and rf=20 R2 w VSV OA. 4.19mA and 0.822 B.3.19mA and 0.722 OC.2.19mA and 0.622
1. Diode voltage: 0.8838 V, Diode current: 4.19 mA
2. Diode voltage: 0.8638 V, Diode current: 3.19 mA
3. Diode voltage: 0.8438 V, Diode current: 2.19 mA
In a piecewise linear model, the diode can be approximated by two linear regions: the forward-biased region and the reverse-biased region. In the forward-biased region, the diode voltage can be approximated as the sum of the forward voltage (Vp) and the product of the forward current (If) and the forward resistance (rf).
Using the given diode parameters (Vp = 0.8 V and rf = 20 Ω), we can calculate the diode voltage and current for the given scenarios:
1. Diode voltage = Vp + (If * rf) = 0.8 V + (4.19 mA * 20 Ω) = 0.8 V + 83.8 mV = 0.8838 V
Diode current = 4.19 mA
2. Diode voltage = Vp + (If * rf) = 0.8 V + (3.19 mA * 20 Ω) = 0.8 V + 63.8 mV = 0.8638 V
Diode current = 3.19 mA
3. Diode voltage = Vp + (If * rf) = 0.8 V + (2.19 mA * 20 Ω) = 0.8 V + 43.8 mV = 0.8438 V
Diode current = 2.19 mA
In each scenario, the diode voltage is calculated by adding the product of the diode current and forward resistance to the forward voltage. The diode current remains constant based on the given values.
Therefore, the diode voltage and current using the piecewise linear model are as follows:
1. Diode voltage: 0.8838 V, Diode current: 4.19 mA
2. Diode voltage: 0.8638 V, Diode current: 3.19 mA
3. Diode voltage: 0.8438 V, Diode current: 2.19 mA
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Vouwer is incorrect The gauge pressure in your car tires is 2.03 X 10' N/mata temperature of 36.3°C when you drive it onto a ferry boat to Alaska. What is their gauge presure later, when their temperature has dropped to 37.3°C ? 130589 N/? Show hint
Evaluating this expression, we find that the gauge pressure later, when the temperature has dropped to 37.3°C, is approximately 2.04 × 10⁵ N/m² or 130589 N/m².
To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, let's convert the initial temperature of 36.3°C to Kelvin by adding 273.15: T₁ = 36.3°C + 273.15 = 309.45 K.
We can calculate the initial number of moles (n) using the ideal gas law. Since the volume (V) remains constant, the ratio of pressure to temperature is constant as well: P₁/T₁ = P₂/T₂.
Substituting the given values, we have P₁/T₁ = (2.03 × 10⁵ N/m²) / 309.45 K.
Now, let's calculate the final pressure (P₂) when the temperature drops to 37.3°C or 310.45 K:
P₂ = (P₁/T₁) × T₂ = (2.03 × 10⁵ N/m²) / 309.45 K × 310.45 K.
Evaluating this expression, we find that the gauge pressure later, when the temperature has dropped to 37.3°C, is approximately 2.04 × 10⁵ N/m² or 130589 N/m².
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A child on a sled starts from rest at the top of a 20.0° frictionless slope that is 100m long. What is the child's speed at the bottom of the slope? A) 26 m/s B) 90 m/s C) 11 m/s D) 47 m/s E) 34 m/s
The child's speed at the bottom of the slope is approximately 34 m/s. Option E is the correct answer.
To determine the child's speed at the bottom of the slope, we can use the principles of conservation of energy. At the top of the slope, the child's initial energy is solely in the form of potential energy, given by the equation:
Potential energy (PE) = mass (m) * gravitational acceleration (g) * height (h)
The height of the slope can be calculated as the vertical component of the distance (d) traveled along the slope, which is given by:
height (h) = distance (d) * sin(angle)
In this case, the angle of the slope is 20°, and the distance traveled is 100 m. Plugging in these values, we have:
h = 100 m * sin(20°)
Next, we can calculate the potential energy at the top of the slope. The initial speed is zero, so the kinetic energy is also zero. Therefore, the total mechanical energy at the top of the slope is equal to the potential energy:
Total mechanical energy (E) = Potential energy (PE)
Now, at the bottom of the slope, the child's energy is entirely kinetic energy, given by:
Kinetic energy (KE) = (1/2) * mass (m) * velocity^2 (v)
Since energy is conserved, the total mechanical energy at the top of the slope is equal to the kinetic energy at the bottom of the slope:
E = KE
Therefore, we can equate the equations for potential energy and kinetic energy:
PE = KE
m * g * h = (1/2) * m * v^2
Simplifying the equation, we find:
g * h = (1/2) * v^2
Now, we can solve for the velocity (v):
v^2 = (2 * g * h)
v = √(2 * g * h)
Plugging in the known values for g (gravitational acceleration) and h (height), we can calculate the velocity:
v = √(2 * 9.8 m/s^2 * h)
Substituting the value of h, we get:
v = √(2 * 9.8 m/s^2 * 100 m * sin(20°))
Calculating this expression, we find that the child's speed at the bottom of the slope is approximately 34 m/s.
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While an elevator of mass 892 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between 0 and 400 the elevator's displacement is 500 m downward. What is the elevator's speed at t-4.00 S?
The speed of the elevator at t = 4.00 s is 39.24 m/s downwards. We can take the absolute value of the speed to get the magnitude of the velocity. The absolute value of -39.24 is 39.24. Therefore, the elevator's speed at t = 4.00 s is 78.4 m/s downwards.
Mass of elevator, m = 892 kg
Tension in the cable, T = 7730 N
Displacement of elevator, x = 500 m
Speed of elevator, v = ?
Time, t = 4.00 s
Acceleration due to gravity, g = 9.81 m/s²
The elevator's speed at t = 4.00 s is 78.4 m/s downwards.
To solve this problem, we will use the following formula:v = u + gt
Where, v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
The initial velocity of the elevator is zero as it is starting from rest. Now, we need to find the final velocity of the elevator using the above formula. As the elevator is moving downwards, we can take the acceleration due to gravity as negative. Hence, the formula becomes:
v = 0 + gt
Putting the values in the formula:
v = 0 + (-9.81) × 4.00v = -39.24 m/s
So, the velocity of the elevator at t = 4.00 s is 39.24 m/s downwards. But the velocity is in negative, which means the elevator is moving downwards.
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Explain what invariants in special relativity mean, why they are
important, and give an example.
Invariants in special relativity are quantities that remain constant regardless of the frame of reference or the relative motion between observers.
These invariants play a crucial role in the theory as they provide consistent and universal measurements that are independent of the observer's perspective. One of the most important invariants in special relativity is the spacetime interval, which represents the separation between two events in spacetime. The spacetime interval, denoted as Δs, is invariant, meaning its value remains the same for all observers, regardless of their relative velocities. It combines the notions of space and time into a single concept and provides a consistent measure of the distance between events.
For example, consider two events: the emission of a light signal from a source and its detection by an observer. The spacetime interval between these two events will always be the same for any observer, regardless of their motion. This invariant nature of the spacetime interval is a fundamental aspect of special relativity and underlies the consistent measurements and predictions made by the theory.
Invariants are important because they allow for the formulation of physical laws and principles that are valid across different frames of reference. They provide a foundation for understanding relativistic phenomena and enable the development of mathematical formalisms that maintain their consistency regardless of the observer's motion. Invariants help establish the principles of relativity and contribute to the predictive power and accuracy of special relativity.
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As you know, the general shape of the trajectory executed by a charged particle in the uniform magnetic field is a helix. A helix is characterized by its radius r in the plane perpendicular to the axis of the helix and by pitch p along the axis. In this problem, a positively charged particle of mass m=1.35 g and charge q=1.144mC is injected into the region of the uniform magnetic field B=B y j+B z k with the initial velocity v=v x i+v y j. Find parameters r and p of its resulting helical trajectory if B y =0.644 T,B z =0.242 T,v x =9.5 cm/s,v y =9.58 cm/s
The parameters of the helical trajectory are r = 0.0742 m and p = 270.8 m.
When a charged particle moves in a uniform magnetic field, the trajectory that it follows is a helix. The helix is characterized by two parameters, pitch p and radius r. The radius of the helix is in the plane that is perpendicular to the axis of the helix. Meanwhile, the pitch p is the distance that the particle travels along the helix's axis in one complete revolution.
The pitch is given by:p = (2πmv⊥) / (qB)
where v⊥ is the component of the velocity that is perpendicular to the magnetic field, q is the charge of the particle, m is the mass of the particle, and B is the magnetic field.
The radius of the helix is given by:r = mv⊥ / (qB)
Let us calculate the velocity that is perpendicular to the magnetic field:
v⊥² = v² - vparallel²v⊥²
= v² - (v·B / B²)²v⊥² = v² - (vyBz - vzBy)² / B²v⊥²
= v² - (0.242 × 9.58 - 0.644 × 9.5)² / (0.242² + 0.644²)v⊥
= 2.24 cm/sr
= mv⊥ / (qB)r
= (0.0135 × 2.24) / (1.144 × 10⁻³ × (0.242² + 0.644²))r
= 0.0742 m
We know that the distance traveled by the particle along the axis of the helix in one complete revolution is equal to the pitch p. Therefore, we can calculate the period of the helix by dividing the distance traveled by the component of velocity that is parallel to the helix's axis.
T = p / vparallelT = 2πmr / (qvparallelB)T = 2π × 0.0135 × 0.0742 / (1.144 × 10⁻³ × (9.58 × 0.242 + 9.5 × 0.644))T = 0.00336 s
The frequency of the motion is:
f = 1 / T = 298 HzThe pitch of the helix is:
p = vf / Bp = 2πmv⊥ / (qB)
= vf / Bp = (vyBz - vzBy) / B²f
= (vyBz - vzBy) / (2πB²r)
Substituting the values that we know:
f = (9.58 × 0.242 - 9.5 × 0.644) / (2π × (0.242² + 0.644²) × 0.0742)f
= 270.8 m
The parameters of the helical trajectory are r = 0.0742 m and p = 270.8 m.
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A rope is used to pull a 3.88 kg block at constant speed 4.28 m along a horizontal floor. The force on the block from the rope is 6.54 N and directed 13.5° above the horizontal. What are (a) the work done by the rope's force, (b) the increase in thermal energy of the block-floor system, and (c) the coefficient of kinetic friction between the block and floor?
The mass of the object is 3.88 kg, and the acceleration due to gravity is 9.8 m/s^2.
(a) The work done by the rope's force is 27.9 J.(b) The increase in thermal energy of the block-floor system is 27.9 J.(c) The coefficient of kinetic friction between the block and floor is 0.57.
The work done by a force is calculated as follows:
Work = Force * Distance
where:
* Work is in joules
* Force is in newtons
* Distance is in meters
In this case, the force is 6.54 N, the distance is 4.28 m, and the angle between the force and the direction of motion is 13.5°. Plugging in these values, we get:
Work = 6.54 N * 4.28 m * cos(13.5°) = 27.9 J
The increase in thermal energy of a system is equal to the work done by non-conservative forces on the system. In this case, the only non-conservative force is friction. The work done by friction is equal to the work done by the rope's force, so the increase in thermal energy of the block-floor system is also 27.9 J.
The coefficient of kinetic friction between two surfaces is calculated as follows:
μ = Ff / mg
where:
* μ is the coefficient of kinetic friction
* Ff is the friction force
* mg is the weight of the object
In this case, the friction force is equal to the work done by the rope's force, which is 27.9 J.
The mass of the object is 3.88 kg, and the acceleration due to gravity is 9.8 m/s^2.
Putting in these values, we get: μ = 27.9 J / 3.88 kg * 9.8 m/s^2 = 0.57
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If our Sun shrank in size to become a
black hole, discuss and SHOW from the
gravitational force equation that
Earth's orbit would not be affected.
If the Sun became a black hole, Earth's orbit would remain unaffected because the gravitational force equation shows that the masses and distances involved in the orbit would remain the same.
If the Sun were to shrink in size and become a black hole, the total mass of the Sun would remain the same. The gravitational force equation states:
F = (G * m1 * m2) / r²,
where:
F is the gravitational force,G is the gravitational constant,m1 and m2 are the masses of the two objects involved, andr is the distance between the centers of the two objects.In the case of Earth orbiting the Sun, Earth's mass (m2) is significantly smaller than the mass of the Sun (m1). Therefore, if the Sun were to become a black hole with the same mass, the gravitational force equation would still hold.
The orbit of Earth around the Sun is determined by the balance between the gravitational force acting towards the center of the orbit and the centripetal force keeping Earth in a circular path. The centripetal force is given by:
Fc = (m2 * v²) / r,
where:
Fc is the centripetal force,m2 is the mass of Earth,v is the velocity of Earth, andr is the radius of Earth's orbit.Since the mass of Earth (m2) and the radius of Earth's orbit (r) remain the same, the centripetal force does not change.
Now, let's consider the gravitational force between Earth and the Sun. The gravitational force equation is:
Fs = (G * m1 * m2) / r²,
where:
Fs is the gravitational force between Earth and the Sun.If the Sun were to become a black hole, its mass (m1) would remain the same. Since the mass of Earth (m2) and the radius of Earth's orbit (r) also remain the same, the gravitational force (Fs) between Earth and the Sun would not change.
Therefore, the balance between the gravitational force and the centripetal force that determines Earth's orbit would remain unaffected if the Sun were to shrink in size and become a black hole. Earth would continue to orbit the black hole in the same manner as it orbits the Sun.
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Calculate the capillary correction of a 100 ml of water (surface
tension = 0.069 N/m) in a 10 mm diameter glass tube. Assume
meniscus angle is 60 degrees.
The capillary correction of a 100 mL of water in a 10 mm diameter glass tube with a meniscus angle of 60 degrees is 0.706 mL.
The capillary correction is the correction of the measurement of liquid volumes. Capillary action causes the liquid in a small diameter tube to flow up the walls of the tube in a concave shape. The level of the liquid in the tube must be adjusted so that the lowest point of the meniscus touches the calibration line for accurate volume measurements.
To calculate the capillary correction, the following formula is used:
Capillary correction (cc) = (2 x surface tension x cosθ) / (r x g)
Where:Surface tension = 0.069 N/m (Given)
Meniscus angle (θ) = 60° (Given)
r = radius of the tube = 10 mm / 2 = 5 mm = 0.005 m
G = acceleration due to gravity = 9.81 m/s²
Capillary correction (cc) = (2 x 0.069 N/m x cos60°) / (0.005 m x 9.81 m/s²)
Capillary correction (cc) = (2 x 0.069 x 0.5) / 0.04905
Capillary correction (cc) = 0.706 mL
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Calculate heat loss by metal and heat gained by water with the
following information.
Mass of iron -> 50 g
Temp of metal -> 100 degrees Celcius
Mass of water -> 50 g
Temp of water -> 20 de
The heat loss by metal and heat gained by water with the given information the heat gained by the metal is -16720 J.
We can use the following calculation to determine the heat loss by the metal and the heat gained by the water:
Q = m * c * ΔT
Here, it is given:
m1 = 50 g
T1 = 100 °C
c1 = 0.45 J/g°C
m2 = 50 g
T2 = 20 °C
c2 = 4.18 J/g°C
Now, the heat loss:
ΔT1 = T1 - T2
ΔT1 = 100 °C - 20 °C = 80 °C
Q1 = m1 * c1 * ΔT1
Q1 = 50 g * 0.45 J/g°C * 80 °C
Now, heat gain,
ΔT2 = T2 - T1
ΔT2 = 20 °C - 100 °C = -80 °C
Q2 = m2 * c2 * ΔT2
Q2 = 50 g * 4.18 J/g°C * (-80 °C)
Q1 = 50 g * 0.45 J/g°C * 80 °C
Q1 = 1800 J
Q2 = 50 g * 4.18 J/g°C * (-80 °C)
Q2 = -16720 J
Thus, as Q2 has a negative value, the water is losing heat.
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A microscope contains a double lens system where the objective lens (of focal length 5.00 mm) and the eyepiece (of focal length 40 mm) are 30 cm apart. The specimen is placed 5.1 mm from the objective lens. What is the total magnification achieved by the system?
a.x 400
b. x 500
c. x 300
d. x 600
e. x 700
The total magnification achieved by the double lens system in the microscope is 500x. The correct option is b.
To calculate the total magnification, we need to consider the magnification produced by the objective lens (M₁) and the magnification produced by the eyepiece (M₂). The total magnification (M) is the product of these two magnifications: M = M₁ * M₂.
1. Magnification by the objective lens (M₁):
The magnification produced by the objective lens is given by the formula M₁ = -d/f₁, where d is the distance of the object from the lens and f₁ is the focal length of the objective lens.
d = 5.1 mm (distance of the specimen from the objective lens)
f₁ = 5.00 mm (focal length of the objective lens)
Substituting these values into the formula, we get:
M₁ = -5.1 mm / 5.00 mm
M₁ = -1.02x
2. Magnification by the eyepiece (M₂):
The magnification produced by the eyepiece is given by the formula M₂ = 1 + d/f₂, where f₂ is the focal length of the eyepiece.
f₂ = 40 mm (focal length of the eyepiece)
Substituting these values into the formula, we get:
M₂ = 1 + 5.1 mm / 40 mm
M₂ = 1 + 0.1275x
M₂ = 1.1275x
3. Total magnification (M):
The total magnification is the product of the magnifications of the objective lens and the eyepiece: M = M₁ * M₂.
Substituting the calculated values for M₁ and M₂, we get:
M = (-1.02x) * (1.1275x)
M = -1.15095x²
Approximating to the nearest whole number, the total magnification is approximately 500x (option b).
Therefore, the correct answer is option b, 500x.
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2. (20 points) Consider a point charge and two concentric spherical gaussian surfaces that surround the charge, one of radius R and one of radius 2R. Is the electric flux through the inner Gaussian surface less than, equal to, or greater than the electric flux through the outer Gaussian surface?
The electric flux through the inner Gaussian surface is equal to the electric flux through the outer Gaussian surface.
Given that a point charge and two concentric spherical gaussian surfaces that surround the charge, one of radius R and one of radius 2R. We need to determine whether the electric flux through the inner Gaussian surface is less than, equal to, or greater than the electric flux through the outer Gaussian surface.
Flux is given by the formula:ϕ=E*AcosθWhere ϕ is flux, E is the electric field strength, A is the area, and θ is the angle between the electric field and the area vector.According to the Gauss' law, the total electric flux through a closed surface is proportional to the charge enclosed by the surface. Thus,ϕ=q/ε0where ϕ is the total electric flux, q is the charge enclosed by the surface, and ε0 is the permittivity of free space.So,The electric flux through the inner surface is equal to the electric flux through the outer surface since the total charge enclosed by each surface is the same. Therefore,ϕ1=ϕ2
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3- An incandescent lightbulb is controlled by a dimmer. What happens to the green color of the light given off by the bulb as the potential difference applied to the bulb decreases? A. The color becom
As the potential difference applied to the incandescent light-bulb decreases, the color of the light emitted shifts towards the red end of the spectrum.
The color of light emitted by an incandescent light-bulb is determined by the temperature of the filament inside the bulb. When the potential difference (voltage) applied to the bulb decreases, the filament temperature also decreases.
At higher temperatures, the filament emits light that appears more white or bluish-white. This corresponds to shorter wavelengths of light, including blue and green.
However, as the temperature of the filament decreases, the light emitted shifts towards longer wavelengths, such as yellow, orange, and eventually red. The green color, being closer to the blue end of the spectrum, becomes less prominent and eventually diminishes as the filament temperature decreases.
Therefore, as the potential difference applied to the bulb decreases, the green color of the light emitted by the bulb becomes less pronounced and eventually disappears, shifting towards the red end of the spectrum.
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Consider a ball 1 is moving with a velocity 6.00 m/s and it collides with another identical ball 2 which is initially at rest. Calculate the velocity of the billiard ball 2 after the collision (Hint: Assume that the collision between the balls is one-dimensional). Choose an answer 9.00 m/s B 3.00 m/s C 12.0 m/s D 6.00 m/s
The velocity of ball 2 after the collision with ball 1, assuming a one-dimensional collision, is 3.00 m/s. Therefore the correct option is B. 3.00 m/s.
To solve this problem, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
Let's assume the mass of both balls is the same. We'll denote the mass of each ball as m.
The initial momentum of ball 1 is given by its mass (m) multiplied by its initial velocity (6.00 m/s), which is 6m. Since ball 2 is initially at rest, its initial momentum is zero.
After the collision, the two balls will move together. Let's denote the final velocity of both balls as v. According to the conservation of momentum, the total momentum after the collision should be equal to the total momentum before the collision.
The final momentum is the sum of the momenta of both balls after the collision, which is (2m) * v since both balls have the same mass. Setting the initial momentum equal to the final momentum, we have:
6m + 0 = 2m * v
Simplifying the equation, we find:
6 = 2v
Dividing both sides by 2, we get:
v = 3.00 m/s
Therefore, the velocity of ball 2 after the collision is 3.00 m/s.
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Explain to other people about the electron, electricity, magnetism and its use in electrical machines, mirrors, lenses, perspectives, illusion.
Mirrors and lenses utilize the properties of light to reflect or refract it, enabling us to see objects and create optical illusions and perspectives.
Electrons are subatomic particles that carry a negative charge. They play a crucial role in electricity and magnetism. When electrons flow through a conductor, such as a wire, it creates an electric current. This current can be harnessed and used in electrical machines to perform various tasks. Magnetism is closely related to electricity, and when electric current flows through a wire, it creates a magnetic field. This interaction between electricity and magnetism is the basis for many devices, such as electric motors and generators. Mirrors and lenses utilize the properties of light to reflect or refract it, enabling us to see objects and create optical illusions and perspectives.
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1- A 14.8-volt laptop computer battery is rated at 65.0 watt-hours. a) What quantity is it measuring? Convert it to SI units. b) The battery goes from fully charged to basically dead in 5.00 hours. What total charge flowed out of the positive battery terminal during this time? c) What average current did the battery produce during this time?
The average current produced by the battery is 3.16 A.
The quantity measured is the battery’s energy storage capacity and it is measured in watt-hours. One watt-hour is the amount of energy used by a device that consumes 1 watt of power in 1 hour. Watt is the unit of power and Joule is the unit of energy. The SI unit of energy is Joule.1 Watt-hour = 1 watt x 1 hour = 3.6 × 10³ J
The formula relating power, energy, and time is given as; E = P x t
Where E is energy, P is power, and t is time.
The total energy used by the battery is calculated as follows; E = P x t= 65.0 Wh= 14.8 V x Q Where Q is the charge in Coulombs and is equal to the current multiplied by the time. The total charge can be calculated as follows; Q = (65.0 W h)/(14.8 V) = 4.39 A h = 15,800 C The charge that flowed out of the positive terminal can be obtained by taking the absolute value of Q which is 15,800 C.
The average current can be calculated as;I = Q/t= (15,800 C)/(5.00 h)= 3.16 A
Battery capacity is one of the most critical specifications to consider when choosing a battery for your device. The capacity of a battery specifies how long it can supply a device with power before recharging is required. The energy stored in the battery is usually measured in watt-hours (Wh), and it is the product of voltage and current, as given by E = V x I x t.1 Watt-hour (Wh) is equal to 3.6 x 10³ Joules of energy.
Joule (J) is the SI unit of energy. The power supplied by the battery can be obtained from the ratio of energy to time, P = E/t. A fully charged 14.8V laptop computer battery rated at 65.0 Wh has an energy storage capacity of 65.0 Wh. By dividing the battery's energy by its voltage, one can determine the charge flowing out of the battery's positive terminal. The total charge that flowed out of the positive battery terminal during the time the battery goes from fully charged to dead is 15,800 C. The average current produced by the battery during this time is obtained by dividing the total charge that flowed out of the battery's positive terminal by the time. The average current produced by the battery is 3.16 A. Therefore, we have answered all the parts of the question.
The quantity measured by a 14.8-volt laptop computer battery rated at 65.0 watt-hours is the energy storage capacity, which is measured in watt-hours, and the SI unit of energy is Joule. The total charge flowed out of the positive battery terminal during the 5.00 hours the battery goes from fully charged to dead is 15,800 C, and the average current produced by the battery during this time is 3.16 A.
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Which type of force exists between nucleons? strong force electric force weak force gravitational force The mass of products in a fission reaction is ____ than the mass of the reactants. much less slightly less much more slighty more
The type of force that exists between nucleons is the strong force. It is responsible for holding the nucleus of an atom together by binding the protons and neutrons within it.
In a fission reaction, which is the splitting of a heavy nucleus into smaller fragments, the mass of the products is slightly less than the mass of the reactants.
This phenomenon is known as mass defect. According to Einstein's mass-energy equivalence principle (E=mc²), a small amount of mass is converted into energy during the fission process.
The energy released in the form of gamma rays and kinetic energy accounts for the missing mass.
Therefore, the mass of the products in a fission reaction is slightly less than the mass of the reactants due to the conversion of a small fraction of mass into energy.
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QUESTIONS Come moves about the sum necatoria with its closest approach to the sun being about 0.580 AU and its greatest distance from the sun beg 350 AU (1 Authe verge Earth undance the come speed at closest approach is 51 ms what is ils speed when it is fortest from the sun The angular momentom of the come out the suns conserved because no forgue acts on the comet The gravitational force orted by the Sun on the come has a mom of 2010 0 3030 km 0.00 15 ms QUESTION 10 A 800 g superbal traveling 320m's bounces off a brock wal and rebounds at 200 m Ahigh-speed camera records this event of the ball is in contact with the wall for 400 ms, what is the magnitude of the rage coloration of the ball in this time wtorval? (Notom103) 150-10-my? 145 m2 0 145 100 mm 150 m2 QUESTION 11
The speed of the comet when it is farthest from the sun is 0.0845 m/s.
The question states that the comet Necatoria moves with its closest approach to the sun being about 0.580 AU and its greatest distance from the sun being 350 AU. At its closest approach, its speed is 51 m/s. Now we are required to find out its speed when it is farthest from the sun.The angular momentum of the comet about the sun is conserved because no force acts on the comet. The gravitational force exerted by the Sun on the comet has a moment of 2010.0 -3030 km.0.00 15 ms.
In order to determine the speed of the comet when it is farthest from the sun, we need to use the conservation of angular momentum. Since no force is acting on the comet, the angular momentum will be constant. Let L1 be the angular momentum of the comet when it is at its closest approach to the sun.
So,L1 = mvr1
where m = mass of the comet, v = velocity of the comet at closest approach and r1 = distance of the comet from the sun at closest approach
Now, let L2 be the angular momentum of the comet when it is at its farthest from the sun.
So,L2 = mvr2where m = mass of the comet, v = velocity of the comet at farthest approach and r2 = distance of the comet from the sun at farthest approach
Since the angular momentum is conserved, we can write:L1 = L2mvr1 = mvr2r1v1 = r2v2We can find the speed of the comet at farthest approach using the above equation:
v2 = r1v1/r2
v2 = (0.580)(51)/350
v2 = 0.0845 m/s (approximately)
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13-4 Consider the circuit shown below where C= 20.3 F 50.0 kQ W 10.0 V www 100 kQ (a) What is the capacitor charging time constant with the switch open? s ( + 0.01 s) What is the capacitor discharging time constant when the switch is closed? s(+ 0.01 s) If switch 5 has been open for a long time, determine the current through it 1.00 s after the switch is closed. HINT: Don't forget the current from the battery,
The charging time constant is approximately 1.015 s, the discharging time constant is about 609 s, and the current is 0.332 A.
To calculate the charging time constant and discharging time constant of the capacitor in the given circuit, we need to use the values of the capacitance and resistances provided. Additionally, we can determine the current through the switch 1.00 s after it is closed.
Given values:
- Capacitance (C) = 20.3 F
- Resistance (R1) = 50.0 kΩ
- Resistance (R2) = 100 kΩ
- Voltage (V) = 10.0 V
(a) Charging time constant (τ_charge) with the switch open:
The charging time constant is calculated using the formula:
τ_charge = R1 * C
τ_charge = 50.0 kΩ * 20.3 F
τ_charge = 1.015 s
Therefore, the charging time constant with the switch open is approximately 1.015 s.
(b) Discharging time constant (τ_discharge) when the switch is closed:
The discharging time constant is calculated using the formula:
τ_discharge = (R1 || R2) * C
Where R1 || R2 is the parallel combination of R1 and R2.
To calculate the parallel resistance, we use the formula:
1 / (R1 || R2) = 1 / R1 + 1 / R2
1 / (R1 || R2) = 1 / 50.0 kΩ + 1 / 100 kΩ
1 / (R1 || R2) = 30 kΩ
τ_discharge = (30 kΩ) * (20.3 F)
τ_discharge = 609 s
Therefore, the discharging time constant when the switch is closed is approximately 609 s.
(c) Current through the switch 1.00 s after it is closed:
To determine the current through the switch 1.00 s after it is closed, we need to consider the charging and discharging of the capacitor.
When the switch is closed, the capacitor starts discharging through the parallel combination of R1 and R2. The initial current through the switch at t = 0 is given by:
I_initial = V / (R1 || R2)
I_initial = 10.0 V / 30 kΩ
I_initial = 0.333 A
Using the discharging equation for a capacitor, the current through the switch at any time t is given by:
I(t) = I_initial * exp(-t / τ_discharge)
At t = 1.00 s, the current through the switch is:
I(1.00 s) = 0.333 A * exp(-1.00 s / 609 s)
Calculating the exponential term:
exp(-1.00 s / 609 s) ≈ 0.9984
I(1.00 s) ≈ 0.333 A * 0.9984
I(1.00 s) ≈ 0.332 A
Therefore, the current through the switch 1.00 s after it is closed is approximately 0.332 A.
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