A 2,000-kg car moving east at 10.0 m/s collides with a 3,000-kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 48.0° north of east and at a speed of 5.98 m/s. Find the speed of the 3,000-kg car before the collision.
To find the speed of the 3,000-kg car before the collision, use the principle of conservation of momentum. Calculate the total momentum of the combined cars after the collision based on the mass and speed. Apply the principle of conservation of momentum again to determine the initial speed of the 3,000-kg car.
Explanation:To find the speed of the 3,000-kg car before the collision, we can use the principle of conservation of momentum. In this case, the momentum before the collision is equal to the momentum after the collision. So, we can equate the momentum of the 2,000-kg car moving east to the momentum of the combined cars moving at an angle of 48.0° north of east.
Using trigonometry, we can determine the components of the velocity of the cars after the collision. The horizontal component will be the speed of the cars moving east, and the vertical component will be the speed of the cars moving north. From the given information, we can calculate the total momentum after the collision based on the mass and speed of the combined cars.
Finally, we can use the principle of conservation of momentum again to determine the speed of the 3,000-kg car before the collision. Rearranging the equation and substituting the known values, we can solve for the initial speed of the 3,000-kg car.
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The speed of the 3,000-kg car before the collision is 7.93 m/s.
Explanation:To find the speed of the 3,000-kg car before the collision, we can use the principle of conservation of momentum. Before the collision, the 2,000-kg car is moving east at 10.0 m/s and the 3,000-kg car is moving north. After the collision, the two cars stick together and move as a unit. Using the given information, we can set up an equation
(2,000 kg)(10.0 m/s) + (3,000 kg)(v) = (5,000 kg)(5.98 m/s
Solving for v, we find that the speed of the 3,000-kg car before the collision is 7.93 m/s.
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A square loop of wire with sides 0.31 m long is placed with its plane perpendicular to a 5.9 T magnetic field. What is the magnetic flux through the loop, in units of Webers?
Answer:
0.567 Weber
Explanation:
side, a = 0.31 m, B = 5.9 T
Area, A = (0.31)^2 = 0.0961 m^2
Magnetic flux, ∅ = B x A
∅ = 5.9 x 0.0961 = 0.567 Weber
A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.97 x 10^4 s. What is the speed at which the satellite travels?
Final answer:
The speed of a satellite in circular orbit is calculated using the period and the radius of the orbit. Without the average altitude of the satellite's orbit, we cannot provide a numerical speed but can describe the method to find it.
Explanation:
To calculate the speed at which a satellite travels in a circular orbit around the Earth, we can use the satellite's period, which is the time it takes to complete one orbit. The formula relates the orbital speed (v), the radius of the orbit (r), and the period (T). Since we are given the mass of Earth (ME) and the period (T), we can calculate the radius of the orbit using the equation for gravitational force, F = G*ME*m/r^2, and centripetal force, F = m*v^2/r. Once we have the radius, we can determine the speed by using the formula v = 2πr/T.
Unfortunately, we're missing the average altitude (therefore the radius) of the satellite's orbit necessary to calculate the speed. Without that information, we can only outline the method to find the speed as described above but cannot provide a numerical answer.
A 5.40 uF parallel-plate, air capacitor has a plate separation of 3.50 mm and is charged to a potential difference of 480 V. Calculate the energy density the region between the plates.
Answer:
Energy density = 0.0831 J/m³
Explanation:
Energy density of a capacitor is given by the expression
[tex]u=\frac{1}{2}\epsilon E^2[/tex]
We have electric field ,
[tex]E=\frac{V}{d}=\frac{480}{3.5\times 10^{-3}}=1.37\times 10^5V/m[/tex]
Here there is no dielectric
So energy density,
[tex]u=\frac{1}{2}\epsilon_0 E^2=\frac{1}{2}\times 8.85\times 10^{-12} \times (1.37\times 10^5)^2=8.31\times 10^{-2}=0.0831 J/m^3[/tex]
Energy density = 0.0831 J/m³
On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86°F (30°C). Determine the maximum κ of the heat pump.
Final answer:
The maximum coefficient of performance (COP) of the heat pump absorbing heat from the outside air at -10°C and transferring it into a home at 30°C is 0.75.
Explanation:
The coefficient of performance (COP) of a heat pump is a measure of its efficiency and is defined as the ratio of the heat delivered to the heat absorbed. In this case, we have a heat pump that absorbs heat from the outside air at -10°C and transfers it into a home at 30°C. To determine the maximum COP of the heat pump, we can use the formula:
COP = T_h / (T_h - T_c)
where T_h is the temperature of the hot reservoir (in this case, 30°C) and T_c is the temperature of the cold reservoir (in this case, -10°C).
Plugging in the values, we get:
COP = 30 / (30 - (-10)) = 30 / 40 = 0.75
Therefore, the maximum COP of the heat pump is 0.75.
An Earth satellite moves in a circular orbit 561 km above Earth's surface with a period of 95.68 min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?
Explanation:
It is given that,
Radius of earth, r = 6371 km
An earth satellite moves in a circular orbit above the Earth's surface, d = 561 km
So, radius of satellite, R = 6371 km + 561 km = 6932 × 10³ m
Time taken, t = 95.68 min = 5740.8 sec
(a) Speed of the satellite is given by :
[tex]v=\dfrac{d}{t}[/tex]
d = distance covered
For circular path, d = 2πR
[tex]v=\dfrac{2\pi \times 6932\times 10^3\ m}{5740.8\ sec}[/tex]
v = 7586.92 m/s
(b) Centripetal acceleration is given by :
[tex]a=\dfrac{v^2}{R}[/tex]
[tex]a=\dfrac{(7586.92\ m/s)^2}{6932\times 10^3\ m}[/tex]
[tex]a=8.3\ m/s^2[/tex]
Hence, this is the required solution.
a) The speed of the satellite is approximately 7584.38 m/s
b) The magnitude of the centripetal acceleration is approximately 8.29 m/s².
To solve this problem, we will calculate both the speed and the magnitude of the centripetal acceleration of the satellite.
(a) Speed of the Satellite
First, we need to determine the radius of the satellite's orbit. The radius of the Earth is R = 6378 km. Since the satellite orbits 561 km above the surface, the radius of the orbit (r) is:
[tex]r = 6378 km + 561 km = 6939 km = 6939000 m[/tex]
The period (T) of the satellite is 95.68 minutes, which we convert to seconds:
[tex]T = 95.68 min \times 60 s/min = 5740.8 s[/tex]
Using the formula for the speed (v) of an object in circular motion:
[tex]v = 2\pi r/T[/tex]
We can plug in the values:
[tex]v = 2\pi \times 6939000 m / 5740.8 s = 7584.38 m/s[/tex]
(b) Magnitude of the Centripetal Acceleration
The centripetal acceleration (ac) is given by:
[tex]a_c = v^2/r[/tex]
Substituting the values we have:
[tex]a_c = (7584.38 m/s)^2 / 6939000 m = 8.29 m/s^2[/tex]
A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is 670 m/s. The gun is pointed directly at the center of the bull's-eye, but the bullet strikes the target 0.025 m below the center. What is the horizontal distance between the end of the rifle and the bull's-eye?
Answer:
47.86 m
Explanation:
Consider the motion along the vertical direction and assume the motion in downward direction as negative
y = vertical displacement = - 0.025 m
a = acceleration = - 9.8 m/s²
t = time taken to strike the target below the center
[tex]v_{oy}[/tex] = initial velocity along the vertical direction = 0 m/s
Using the kinematics equation
y = [tex]v_{oy}[/tex] t + (0.5) a t²
Inserting the values
- 0.025 = (0) t + (0.5) (- 9.8) t²
t = 0.0714 sec
Consider the motion along the horizontal direction
x = horizontal displacement = horizontal distance between the end of the rifle and the bull's-eye
v₀ = velocity along the horizontal direction = 670 m/s
Since there is no acceleration along the horizontal direction, we have
x = v₀ t
inserting the values
x = (670) (0.0714)
x = 47.86 m
The horizontal distance between the end of the rifle and the bull's-eye is approximately 47.9 meters.
To solve this problem, we need to determine the horizontal distance from the rifle to the bull's-eye based on the given data.
Understand the Problem: The bullet is fired horizontally with a muzzle speed of 670 m/s and hits the target 0.025 m below the center due to gravity.
Use Kinematics: Gravity affects the vertical motion, and the horizontal motion is constant because there's no acceleration horizontally in the given conditions.
Calculate the Time it Takes for the Bullet to Drop 0.025 m:
The vertical distance (y) the bullet drops can be calculated using the equation of motion under gravity:
[tex]y = \frac{1}{2} g t^2[/tex]
Where:
[tex]y[/tex] is the vertical drop (0.025 m)[tex]g[/tex] is the acceleration due to gravity (approximately 9.8 m/s²)[tex]t[/tex] is the time in secondsRearranging for [tex]t[/tex]:
[tex]t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2 \times 0.025}{9.8}} \approx 0.0714 \text{ seconds}[/tex]
Calculate the Horizontal Distance (x):
Now that we have the time [tex]t[/tex], we can find the horizontal distance using the constant horizontal velocity.
[tex]x = v \cdot t[/tex]
Where:
[tex]v[/tex] is the muzzle speed (670 m/s)[tex]t[/tex] is the time calculated above (0.0714 seconds)[tex]x = 670 \times 0.0714 \approx 47.9 \text{ meters}[/tex]
Car A uses tires for which the coefficient of static friction is 0.169 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 23.7 m/s. Car B uses tires for which the coefficient of static friction is 0.826 on the same curve. What is the maximum speed at which car B can negotiate the curve?
Answer:
[tex]v_B = 52.4 m/s[/tex]
Explanation:
For unbanked road the maximum friction force will provide centripetal force to the car.
So here we will have
[tex]F_c = \frac{mv^2}{R}[/tex]
Since we know that centripetal force here is due to friction force
[tex]F_c = F_f[/tex]
[tex]\mu mg = \frac{mv^2}{R}[/tex]
now for two cars we will have
[tex]\mu_A m_A g = \frac{m_A v_A^2}{R}[/tex]
also we have
[tex]\mu_B m_B g = \frac{m_B v_B^2}{R}[/tex]
now by division of two equations
[tex]\frac{\mu_A}{\mu_B} = \frac{v_A^2}{v_B^2}[/tex]
[tex]\frac{0.169}{0.826} = \frac{23.7^2}{v_B^2}[/tex]
so we will have
[tex]v_B = 52.4 m/s[/tex]
Final answer:
The maximum speed at which car B can negotiate the curve is approximately 52.4 m/s, assuming the same curve and conditions as for car A.
Explanation:
To determine the maximum speed at which car B can negotiate the curve, we need to use the relationship between the coefficient of static friction and the maximum speed a car can maintain without slipping on a curve.
The centripetal force required to keep a car moving in a circle of radius r is provided by the frictional force, which is the product of the coefficient of static friction μ and the normal force N, which for a flat curve is equal to the weight of the car.
Since car A, with a coefficient of static friction of 0.169, can negotiate the curve at a speed of 23.7 m/s, we can use proportionality to find the speed for car B, as the friction forces are proportional to the coefficients of static friction of their respective tires:
VB = VA ∙ √(μB / μA)
Substituting the given values:
VB = 23.7 m/s ∙ √(0.826 / 0.169)
VB = 23.7 m/s ∙ √(4.89) ≈ 23.7 m/s ∙ 2.21
VB ≈ 52.4 m/s
Therefore, the maximum speed at which car B can negotiate the curve is approximately 52.4 m/s, assuming the same curve and conditions as for car A.
Saturated steam coming off the turbine of a steam power plant at 40°C condenses on the outside of a 3-cm-outerdiameter, 35-m-long tube at a rate of 63 kg/h. Determine the rate of heat transfer from the steam to the cooling water flowing through the pipe. The properties of water at the saturation temperature of 40°C are hfg
Answer:
Q = 30.07 kJ/sec
Explanation:
GIVEN DATA:
temperature of steam = 40 degree celcius
mass flow rate = 63 kg/h
from saturated water tables,
from temperature 40 °, enthalpy of evaporation[tex]h_f[/tex] value is 2406 kj/kg
rate of heat transfer (Q) can be determine by using following relation
[tex]Q = \dot{m} h_f[/tex]
putting all value to get Q value
Q = 45 *2406
Q = 108270 kJ/h
[tex]Q = 108270 *\frac{1}{3600} kJ/s[/tex]
Q = 30.07 kJ/sec
Explanation:
According to the water table, the value of enthalpy of evaporation at a temperature of [tex]40^{o}C[/tex] is 2406.0 kJ/kg.
Hence, we will calculate the rate of heat transfer by using the formula as follows.
Q = [tex]m \times h_{fg}[/tex]
where, m = mass
[tex]h_{fg}[/tex] = enthalpy of evaporation
Putting the given values into the above formula as follows.
Q = [tex]m \times h_{fg}[/tex]
= [tex]63 kg/h \times 2406.0 kJ/kg[/tex]
= 151578 kJ/h
or, = [tex]151578 \times \frac{1}{3600} kJ/s[/tex]
= 42.105 kW
Thus, we can conclude that rate of heat transfer from the steam to the cooling water flowing through the pipe is 42.105 kW.
What's the cosmological principle?
Answer:
it is concept that considered all matter in universe homogeneous and it is acceptable on large scale.
Explanation:
The cosmological principle is the concept that, when recognized on a large basis, the visual allocation of matter and energy is homogeneous and isotropic, since the forces are expected to act homogeneously across the universe and should therefore not produce any observed errors in the wide-scale. This eliminates any uncertainty about irregularities in the characteristics of matter and even includes the development of this matter since its creation during the Big Bang.
Final answer:
The cosmological principle states that the universe at any given time is the same everywhere on a large scale. This principle is crucial for understanding and studying the universe, as it allows us to make assumptions about its properties based on what we observe locally. It is a fundamental concept in cosmology.
Explanation:
The cosmological principle is the assumption that, on a large scale, the universe at any given time is the same everywhere, meaning it is isotropic and homogeneous. This principle is the starting assumption for nearly all theories that describe the entire universe. It implies that the universe is about the same everywhere, apart from changes with time, and that the part we can see around us is representative of the whole.
A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor with X 24 ohms, what is the circuit impedance? a. 25.70 b. 29.2Q 32.4Q c. d. 35.90 5
Answer:
the impedance of the circuit is 25.7 ohms.
Explanation:
It is given that,
Voltage, V = 50 volts
Frequency, f = 60 Hz
Resistance, R = 25 ohms
Capacitive resistance, [tex]X_C=18\ ohms[/tex]
Inductive resistance, [tex]X_L=24\ ohms[/tex]
We need to find the impedance of the circuit. It is given by :
[tex]Z=\sqrt{R^2+(X_L-X_C)^2}[/tex]
[tex]Z=\sqrt{25^2+(24-18)^2}[/tex]
Z = 25.7 ohms
So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.
Suppose 1.00 L of a gas with \gamma γ = 1.30, initially at 273 K and 1.00 atm, is suddenly compressed adiabatically to half its initial volume. If the gas is then cooled to 273 K at constant pressure, what is its final volume in L?
Answer:0.4061 L
Explanation:
let initial volume of gas be [tex]V_1[/tex]=1 L
[tex]\gamma[/tex] =1.3
initial temprature[tex]\left ( T_1\right )[/tex]=273k
Now if gas is compressed adiabatically to half of its initial volume then
[tex]V_2[/tex]=0.5L
Using adiabatic relation
[tex]TV^{\gamma -1}[/tex]=constant
[tex]T_1V_1^{\gamma -1}[/tex]=[tex]T_2V_2^{\gamma -1}[/tex]
[tex]273\times \left ( \frac{1}{0.5}\right )^\left ( {\gamma -1}\right )[/tex]=[tex]T_2[/tex]
[tex]T_2[/tex]=273[tex]\times \left ( 2\right )^\left ( 0.3\right )[/tex]
[tex]T_2[/tex]=336.1k
Now the is cooled at constant pressure i.e.
[tex]\frac{V_2}{T_2}[/tex]=[tex]\frac{V_3}{T_3}[/tex]
[tex]\frac{0.5}{336.1}[/tex]=[tex]\frac{V_3}{273}[/tex]
[tex]\frac{0.5}{336.1}[/tex]=[tex]\frac{V_3}{273}[/tex]
[tex]V_3[/tex]=0.406 L
The final volume of the gas after it is cooled to 273 K at constant pressure, following an adiabatic compression to half its initial volume, will remain at 0.5 L.
The question involves an adiabatic compression of a gas followed by cooling at constant pressure. To find the final volume after the gas is cooled to 273 K at constant pressure, we can use the ideal gas law before and after the compression, considering that for adiabatic processes, PVγ is constant (where P is pressure, V is volume, and γ is the adiabatic index). After the adiabatic compression, we have a new volume of 0.5 L and a new pressure. We then cool the gas at this new pressure back to the original temperature of 273 K. By the ideal gas law (PV = nRT), at constant temperature and number of moles (n), the final volume (Vf) will be equal to the initial volume (Vi) because the pressure has remained constant after the compression step.
Therefore, even after cooling, the final volume will remain at 0.5 L because pressure does not change during the cooling process at constant pressure.
At what frequency will a 3.0 μF capacitor have a reactance of 7.0 kΩ?
Answer:
Frequency, f = 7.57 Hz
Explanation:
It is given that,
Capacitance, [tex]C=3\ \mu F=3\times 10^{-6}\ F[/tex]
Capacitive reactance, [tex]X_C=7\ k\Omega=7\times 10^3\ \Omega[/tex]
We need to find the frequency. The capacitive reactance of the capacitor is given by :
[tex]X_C=\dfrac{1}{2\pi fC}[/tex]
f is the frequency
[tex]f=\dfrac{1}{2\pi CX_C}[/tex]
[tex]f=\dfrac{1}{2\pi \times 3\times 10^{-6}\ F\times 7\times 10^3\ \Omega}[/tex]
f = 7.57 Hz
Hence, this is the required solution.
Rectangular Loop A current I flows in a rectangular loop of wire placed on the y plane and centered about the origin. The length of the two sides are a and b (a) Find the magnetic field at any location (0,0, 2) on the z-axis. (b) For za, b, what does your expression simplify to? (c) Compare your answer to (b) with the on-axis field for a circular loop at distances much greater than the loop radius Z. Three Parallel Wires Consider three co-planar equally-spaced (by distance d) parallel wires of negligible radius. Each wire carries current I, a in the same direction (a) Where does the magnetic field vanish? (b) Sketch the magnetic field nes in the plane perpendicular to the current flow e) Suppose you displace the middle wire a sinall distance δ in a direction perpendicular to the plane containing the wires and then release it. Analyze and describe the resulting motion of the middle wire
Answer:
mathy math
Explanation:
do you eat at baskin-robbins?
btw, answer is probably 78 (6-9)x 67,587r Xy
A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the equilibrium position. (a) Find the position x of the mass at the times t = π/12, π/8, π/6, π/4, and 9π/32 s. (Use g = 32 ft/s2 for the acceleration due to gravity.)
The problem requires the understanding of simple harmonic motion in physics. The position of an oscillating mass attached to a spring can be calculated using the SHM formulas and given parameters: weight of the mass, stretch of the spring, and acceleration due to gravity. The amplitude, spring constant, angular frequency, and phase constant are figured out before plugging them into the positional formula.
Explanation:This question involves the concept of simple harmonic motion (SHM) in Physics. In this scenario, a mass attached to a spring and released from a certain position will oscillate in SHM. The position of the mass at specific times (t = π/12, π/8, π/6, π/4, and 9π/32 s) can be calculated using the SHM formulas.
Since the mass is released from 6 inches (or 0.5 feet) below the equilibrium position, this value represents the amplitude (A) of the oscillation. Additionally, we know that the spring constant (k) can be found from the weight of the mass and the stretch of the spring. This gives us k = W/d = 20 lbs / 0.5 ft = 40 lbs/ft.
The angular frequency (ω) of the oscillation can be derived from ω = sqrt(k/m), where m is the mass of the object. Using the information provided (considering g = 32 ft/s2), we find m = W/g = 20 lbs / 32 ft/ s2 = 0.625 slugs. Hence, ω = sqrt(40 lbs/ft / 0.625 slugs) = 8 rad/s.
Finally, using the formula for the position in a SHM (x = A*cos(ω*t + φ), where φ is the phase constant), and knowing that the mass was initially released from rest, we can determine that φ = 0. As a result, the position at the specified times can be calculated by substituting the corresponding values into the relation.
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To find the position of the mass at different times, use the equation for simple harmonic motion (SHM) and calculate the angular frequency. Then plug in the values and use the equation x = A * cos(ωt + φ) to find the positions.
Explanation:The position x of the mass at different times can be found using the equation for simple harmonic motion (SHM):
x = A * cos(ωt + φ)
where x is the position, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.
In this case, the amplitude A is 6 inches and the angular frequency ω can be found using the formula ω = sqrt(k/m), where k is the spring constant and m is the mass. The spring constant k can be found using the formula k = (g * m)/A, where g is the acceleration due to gravity.
By plugging in the values and calculating ω, you can then find the positions x at the given times using the equation x = A * cos(ωt + φ).
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A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can jump with the same initial speed at all angles, how much distance does he lose (in m) by taking off at 429?
Answer:
0.02 m
Explanation:
R₁ = initial distance jumped by jumper = 7.4 m
R₂ = final distance jumped by jumper = ?
θ₁ = initial angle of jump = 45°
θ₂ = final angle of jump = 42.9°
[tex]v[/tex] = speed at which jumper jumps at all time
initial distance jumped is given as
[tex]R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}[/tex]
final distance jumped is given as
[tex]R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}[/tex]
Dividing final distance by initial distance
[tex]\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}[/tex]
[tex]\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}[/tex]
[tex]R_{2} =7.38[/tex]
distance lost is given as
d = [tex]R_{1} - R_{2} [/tex]
d = 7.4 - 7.38
d = 0.02 m
Suppose that an experiment determines that the amount of work required for a Force field to move a particle from the point ( 1 , 2 ) to the point ( − 5 , 3 ) along the curve C 1 is 1.2 J and the work done by the field in moving the particle along another curve C 2 between the same two points is 1.4 J. What can you say about the field, and why? g
Answer:
From the given information we can infer that the field is not conservative.
Explanation:
For a conservative field the work done on an object in moving it from a position given by co-ordinates [tex](x_{1},y_{1},z_{1})[/tex] to another position with co-ordinates [tex](x_{2},y_{2},z_{2})[/tex] shall be independent of the path we take in between to reach our final position (by definition of a conservative field). But in the given case since the initial and the final position of both the curves [tex]C_{1},C_{2}[/tex] coincide but the work done along both the paths is different thus we conclude that the field is not conservative.
Consider a generator that rotates its 200 turn, 0.18 m diameter coil at 3250 rpm in a 0.65 T field. Calculate the peak voltage of the generator.
Answer:
1124.8 Volt
Explanation:
N = 200, Diameter = 0.18 m, Radius, r = 0.09 m, f = 3250 rpm = 54.17 rps
B = 0.65 T
the peak value of induced emf is given by
e0 = N x B x A x ω
e0 = 200 x 0.65 x 3.14 x 0.09 x 0.09 x 2 x 3.14 x 54.17
e0 = 1124.8 Volt
The peak voltage of the generator can be calculated using Faraday's law of electromagnetic induction, taking into consideration the number of turns in the coil, the strength of the magnetic field, the area of the coil, and the speed at which the coil is rotating. In this case, it's 1100 volts.
Explanation:The subject of this question is a practical application of Faraday's law of electromagnetic induction, which is a topic typically covered in high school or college-level physics. We can find the peak voltage of a rotating generator using the formula for Induced electromotive force (emf), also known as voltage, in a rotating coil, which is given by ε_max = NBAω. Here, N is the number of turns in the coil, B is the magnetic field strength, A is the area of the coil, and ω is the angular velocity.
Let's calculate each parameter. The area A of the coil can be calculated using the formula for the area A = πr^2, where r is the radius of the coil. Given the diameter d=0.18 m, we get r = d/2 = 0.09 m, which gives A = π * (0.09 m)^2 = 0.025 m^2. The angular velocity ω in rad/s is calculated using the relation ω = 2πf, where f is the frequency in Hz. Given the frequency in rpm (revolutions per minute), we convert it to Hz by dividing by 60, hence f = 3250/60 = 54.16 Hz. Then ω = 2π * 54.16 Hz = 340.53 rad/s. Now, we can substitute the values into the emf equation: ε_max = (200 turns) * (0.65 T) * (0.025 m^2) * (340.53 rad/s) = 1100 volts. Hence, the peak voltage of the generator is 1100 volts.
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A particle with a charge of +4.50 nC is in a uniform electric field E+ directed to the negative x direction. It is released from rest, and after it has moved 8.00 cm , its kinetic energy is found to be 1.50×10^−6 J . You may want to review (Pages 567 - 572) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Parallel plates and conservation of energy.
Part A: What work was done by the electric force?Part B: What was the change in electric potential over the distance that the charge moved?Part C: What is the magnitude of E+ ?Part D: What was the change in potential energy of the charge?
(a) [tex]1.50\cdot 10^{-6}J[/tex]
According to the work-energy theorem, the work done by the electric force is equal to the kinetic energy gained by the particle, so we can write:
[tex]W=K_f - K_i[/tex]
where
W is the work done by the electric force
[tex]K_f[/tex] is the final kinetic energy of the particle
[tex]K_i[/tex] is the initial kinetic energy of the particle
Since the particle starts from rest, [tex]K_i = 0[/tex]. Moreover,
[tex]K_f = 1.50\cdot 10^{-6}J[/tex]
Therefore, the work done by the electric force is
[tex]W=K_f = 1.50\cdot 10^{-6}J[/tex]
(b) -333.3 V
According to the law of conservation of energy, the gain in kinetic energy of the particle must correspond to a loss in electric potential energy, so we can write:
[tex]\Delta K = -\Delta U[/tex]
Where
[tex]\Delta K = 1.50\cdot 10^{-6} J[/tex] is the gain in kinetic energy
[tex]\Delta U[/tex] is the loss in electric potential energy
So we have
[tex]\Delta U = - \Delta K = -1.50\cdot 10^{-6}J[/tex]
The loss in electric potential energy can be rewritten as
[tex]\Delta U = q \Delta V[/tex]
where
[tex]q=+4.50 nC = 4.5\cdot 10^{-9} C[/tex] is the charge of the particle
[tex]\Delta V[/tex] is the change in electric potential over the distance the charge has moved
Solving for [tex]\Delta V[/tex],
[tex]\Delta V= \frac{\Delta U}{q}=\frac{-1.50\cdot 10^{-6}}{4.5\cdot 10^{-9}}=-333.3 V[/tex]
(c) 4166 V/m
The magnitude of the electric field is given by
[tex]E=\frac{|\Delta V|}{d}[/tex]
where
[tex]|\Delta V|[/tex] is the magnitude of the change in electric potential
d is the distance through which the charge has moved
Since we have
[tex]|\Delta V|=333.3 V\\d = 8.00 cm = 0.08 m[/tex]
The magnitude of the electric field is
[tex]E=\frac{333.3}{0.08}=4166 V/m[/tex]
(d) [tex]-1.50\cdot 10^{-6}J[/tex]
The change in electric potential energy of the charge has already been calculated in part (b), and it is
[tex]\Delta U = -1.50\cdot 10^{-6}J[/tex]
What is the minimum number of 100 Ohm resistors that you need to design an effective resistor with 275 Ohm resistance.
Answer:
The minimum number of 100 Ω resistors that i need to design an effective resistor with 275Ω resistance are 8 resistors.
Explanation:
(2 Resistors of 100Ω in parallel) in series with (4 resistors of 100Ω in parallel) in series with 2 resistors of 100Ω.
2 resistors in parallel of 100Ω = 50Ω
+
4 resistors in parallel of 100Ω = 25Ω
+
2 resistors in series of 100 2Ω = 200Ω
=
275Ω
What is the reactance of a capacitor at a 8.0 μF frequency of 56.0 Hz?
Answer:
35.55 ohm
Explanation:
C = 80 micro farad
F = 56 hertz
Reactance
Xc = 1 / (2 × pi × F × C)
Xc = 1 / ( 2 × 3.14 × 56 × 80 × 10^-6)
Xc = 35.55 ohm
What type of mage can never be formed by a converging lens?
Answer: Real image
Explanation:
converging lens will only produce a real image if the object is located beyond the focal point (i.e., more than one focal length away).
A 5-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in the same direction as the jet at a velocity of 10 m/s. The water splatters in all directions in the plane of the plate. How much force does the water stream exert on the plate?
To calculate the force exerted by a stream of water on a moving plate, determine the change in momentum of the water as it hits the plate. By taking into account the relative velocity of the water to the plate, its cross-sectional area, and the density of water, we find the force exerted to be 17670 Newtons.
Explanation:The subject of this question is related to physics, specifically fluid dynamics. It involves calculating the force exerted by a jet of water on a moving plate. The plate and the water jet are moving in the same direction, but at different speeds. To calculate this, we need to understand the change in momentum of the water as it hits the plate.
First, we calculate the velocity of water relative to the plate, which is 40 m/s - 10 m/s = 30 m/s. The water's diameter is 5 cm, which gives us a cross-sectional area (A) of 19.63 cm2. We can obtain the volume flow rate (Q) of the water from Q = Av, which gives 0.589 m3/s. Multiplying this by the density of water (ρ=1000 kg/m3) gives a mass flow rate (ṁ) of 589 kg/s.
The force exerted by the water on the plate is equal to the rate of change of momentum of the water. Since the water is coming to rest on the plate, the initial momentum of the water relative to the plate is -ṁv, and the final momentum is zero. Therefore, the force exerted on the plate is Δp/Δt = ṁv = 589 kg/s * 30 m/s = 17670 N.
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The force exerted by the water stream on the plate is [tex]\( 2356.2 \, \text{N} \).[/tex]
To determine the force exerted by the water stream on the moving plate, we need to apply the principles of fluid dynamics and the conservation of momentum.
Given Data:
Diameter of the water jet, [tex]\( D = 5 \, \text{cm} = 0.05 \, \text{m} \)[/tex]Velocity of the water jet relative to the ground, [tex]\( V_{\text{jet}} = 40 \, \text{m/s} \)[/tex]Velocity of the plate, [tex]\( V_{\text{plate}} = 10 \, \text{m/s} \)[/tex]Step-by-Step Solution:
1. Calculate the cross-sectional area of the jet:
[tex]\[ A = \pi \left(\frac{D}{2}\right)^2 = \pi \left(\frac{0.05}{2}\right)^2 = \pi \left(0.025\right)^2 = \pi \times 0.000625 = 0.001963495 \, \text{m}^2 \][/tex]
2. Determine the velocity of the water relative to the plate:
[tex]\[ V_{\text{rel}} = V_{\text{jet}} - V_{\text{plate}} = 40 \, \text{m/s} - 10 \, \text{m/s} = 30 \, \text{m/s} \][/tex]
3. Calculate the mass flow rate of the water jet:
[tex]\[ \dot{m} = \rho \cdot A \cdot V_{\text{jet}} \][/tex]
Assuming the density of water [tex]\( \rho \) is \( 1000 \, \text{kg/m}^3 \):[/tex]
[tex]\[ \dot{m} = 1000 \, \text{kg/m}^3 \times 0.001963495 \, \text{m}^2 \times 40 \, \text{m/s} \][/tex]
[tex]\[ \dot{m} = 78.54 \, \text{kg/s} \][/tex]
4. Apply the conservation of momentum to find the force:
The change in momentum of the water relative to the plate will result in the force exerted on the plate. Since the water splatters in all directions in the plane of the plate, we assume the relative velocity of the water to the plate goes to zero after hitting the plate (as it splatters and no longer moves in a single direction).
The force exerted by the water on the plate is given by the rate of change of momentum:
[tex]\[ F = \dot{m} \cdot V_{\text{rel}} \][/tex]
Substituting the values:
[tex]\[ F = 78.54 \, \text{kg/s} \times 30 \, \text{m/s} \][/tex]
[tex]\[ F = 2356.2 \, \text{N} \][/tex]
The Earth is 81.25 times as massive as the Moon and the radius of the Earth is 3.668 times the radius of the Moon. If a simple pendulum has a frequency of 2 Hz on Earth, what will be its frequency on the Moon? a) 0.407 Hz b) 0.814 Hz c) 1.34 Hz d) 1.22 Hz e) 1.63 Ha
Answer:
option (b)
Explanation:
Mass of moon = m
Mass of earth = 81.25 x mass of moon = 81.25 m
Radius of moon = r
radius of earth = 3.668 x radius of moon = 3.668 r
Frequency on earth = 2 Hz
Let the frequency on moon is f.
The formula for the frequency is given by
[tex]f = \frac{1}{2\pi }\times \sqrt{\frac{g}{l}}[/tex]
The value of acceleration due to gravity on earth is g.
ge = G Me / Re^2
ge = G x 81.25 m / (3.668 r)^2
ge = 6.039 x G m / r^2 = 6.039 x gm
ge / gm = 6.039
Now use the formula for frequency
[tex]\frac{fe}{fm} = \sqrt{\frac{ge}{gm}}[/tex]
[tex]\frac{2}{fm} = \sqrt{\frac{6.039 gm}{gm}}[/tex]
[tex]\frac{2}{fm} = 2.46[/tex]
fm = 0.814 Hz
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 21.5 ° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 55.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.212. Find the magnitude of the force that the tow bar exerts on the skier.
Mass = 55.0 kg
Coefficient of friction = 0.212
Slope = 21.5 degrees
Gravity = 9.81
Magnitude of force = 55*9.81*(0.212*cos21.5 + sin21.5)
Magnitude of force = 304.17 Newtons
Round the answer as needed.
Answer:
Hope it can help you and mark me as a brainlist..
13) A pendulum of mass M, length L, amplitude A, and on the earth with acceleration due to gravity of g has a period T. How long would the pendulum need to be in order to to have a period of 2T?
Answer:
4L
Explanation:
mass of pendulum = M, length of pendulum, L1 = L,
acceleration due to gravity = g, Amplitude = A
Use the formula for the time period for simple pendulum
[tex]T = 2\pi \sqrt{\frac{L}{g}}[/tex] ..... (11)
Let the new length be L2 for which the time period is 2T.
[tex]2T = 2\pi \sqrt{\frac{L_{2}}{g}}[/tex] ..... (2)
Divide equation (2) by equation (1)
[tex]\frac{2T}{T} = \sqrt{\frac{L_{2}}{L}}[/tex]
So, L2 = 4 L
Thus, the length of the pendulum is 4 L.
A 64.4-kg firefighter climbs a flight of stairs 21.3-m high. How much work (in J) does he do?
Answer:
Work done by firefighter = 13456.57 J
Explanation:
Here work done is equal to potential energy gained by him.
Potential energy, PE = mgh, where m is the mass, g is acceleration due to gravity value and h is the height.
Mass, m = 64.4 kg
Acceleration due to gravity, g = 9.8 m/s²
Height, h = 21.3 m
Substituting
Potential energy, PE = mgh = 64.4 x 9.81 x 21.3 = 13456.57 J
Work done by firefighter = 13456.57 J
Find the de Broglie wavelength of a ball of mass 0.20 kg just before it strikes the Earth after being |dropped from a building 50 m tall. IMPORTANT: Step by step *typed* format preferred. If hand written, please write clearly and legibly. Please be sure answer is CORRECT. Thanks!
Answer:
Wavelength of the ball is [tex]\lambda=1.05\times 10^{-34}\ m[/tex]
Explanation:
It is given that,
Mass of the ball, m = 0.2 kg
It strikes the Earth after being dropped from a building of 50 m tall, h = 50 m
In this time, the potential energy is converted to kinetic energy. On applying the conservation of energy as :
[tex]\dfrac{1}{2}mv^2=mgh[/tex]
[tex]v=\sqrt{2gh}[/tex].............(1)
The De-broglie wavelength of the ball is given by :
[tex]\lambda=\dfrac{h}{mv}[/tex]
[tex]\lambda=\dfrac{h}{m\sqrt{2gh}}[/tex]
[tex]\lambda=\dfrac{6.63\times 10^{-34}\ J-s}{0.2\ kg\times \sqrt{2\times 9.8\ m/s^2\times 50\ m}}[/tex]
[tex]\lambda=1.05\times 10^{-34}\ m[/tex]
Hence, this is the required solution.
The mass of a hot-air balloon and its cargo (not including the air inside) is 170 kg. The air outside is at 10.0°C and 101 kPa. The volume of the balloon is 530 m3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (Air density at 10.0°C is 1.244 kg/m3.)
Answer:
108.37°C
Explanation:
P₁ = Initial pressure = 101 kPa
V₁ = Initial volume = 530 m³
T₁ = Initial temperature = 10°C = 10+273.15 =283.15 K
P₂ = Final pressure = 101 kPa (because it is open to atmosphere)
V₂ = Final volume = 530 m³
P₁V₁ = n₁RT₁
⇒101×530 = n₁RT₁
⇒53530 J = n₁RT₁
P₂V₂ = n₂RT₂
⇒53530 J = n₂RT₂
[tex]\frac{m_1}{m_2}=\frac{\rho V_1}{\rho V_1-170}\\\Rightarrow \frac{m_1}{m_2}=\frac{1.244\times 530}{1.244\times 530-170}=1.347\\\Rightarrow \frac{m_1}{m_2}=1.347\\\Rightarrow \frac{n_1}{n_2}=1.347[/tex]
Dividing the first two equations we get
[tex]1=\frac{n_1}{n_2}\frac{T_1}{T_2}\\\Rightarrow 1=1.347\frac{283.15}{T_2}\\\Rightarrow T_2=1.347\times 283.15= 381.52\ K[/tex]
∴Temperature must the air in the balloon be warmed before the balloon will lift off is 381.25-273.15 = 108.37°C
To lift off, the air in the balloon must be warmed to a temperature of approximately 916.93 K.
Explanation:To calculate the temperature to which the air in the balloon must be warmed before it lifts off, we can use the ideal gas law equation: PV = nRT.
The volume of the balloon is given as 530 m3, the pressure is atmospheric pressure (101 kPa), and the mass is 170 kg (which can be converted to moles using the molar mass of air). We can rearrange the equation to solve for temperature:
T = PV/(nR)
Substituting the values, we have:
T = (101,000 Pa) × (530 m3) / (n × 8.31 J/mol·K)
Where n is the number of moles of air. To find the number of moles, we can use the density of air at 10.0°C (1.244 kg/m3):
n = mass / molar mass = 170 kg / (1.244 kg/m3 × 530 m3)
Substituting this value for n and solving for T:
T = (101,000 Pa) × (530 m3) / ((170 kg / (1.244 kg/m3 × 530 m3)) × 8.31 J/mol·K)
Calculating, we find T ≈ 916.93 K.
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A 43.0 cm diameter wheel accelerates uniformly from 86.0 rpm to 362.0 rpm in 3.1 s. How far (in meters) will a point on the edge of the wheel have traveled in this time? socratic.org
Answer:
15.6 m
Explanation:
d = diameter of the wheel = 43 cm = 0.43 m
r = radius of the wheel = (0.5) d = (0.5) (0.43) = 0.215 m
w₀ = initial angular velocity = 86 rpm = 9.01 rad/s
v₀ = initial linear speed = r w₀ = (0.215) (9.01) = 1.94 m/s
w = final angular velocity = 362 rpm = 37.91 rad/s
v = final linear speed = r w = (0.215) (37.91) = 8.15 m/s
t = time taken = 3.1 s
Using the equation
v = v₀ + at
8.15 = 1.94 + a (3.1)
a = 2 m/s²
d = distance traveled by a point at the edge of the wheel
Using the equation
d = v₀ t + (0.5) at²
d = (1.94) (3.1) + (0.5) (2) (3.1)²
d = 15.6 m