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Leokris [45]
2 years ago
6

For the circuit in the previous part, what happens to the maximum current if the frequency is doubled and the inductance is halv

ed
Physics
1 answer:
tamaranim1 [39]2 years ago
4 0

Answer:

Following are the responses to these question:

Explanation:

Since the max^{m} is the current of ckt which depend on the reactance which   inductor that also enables the ckt and inductor resistance (X_L) for capacities

\to X_{C}=\frac{1}{W L}

for

\to X_{L}=wL

When w \longrightarrow 2w

L\longrightarrow \frac{L}{2}

then

\to X_{L}=2 w \times \frac{L}{2}=wL

 therefore, X_{L} remains at the same so, the maximum current remains the in same ckt.

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Weight = mass * gravity = 60 kg * 3.75 m/s² = 225 N

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A 0.5 kg turntable with a radius of 15 cm is rotating 78 times per minute.
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At an instant when a particle of mass 80 g has a velocity of 25 m/s in the positive y direction, a 75-g particle has a velocity
dalvyx [7]

Answer:

16 m/s

Explanation:

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y momentum = 0.080 *25 = 2

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3 years ago
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It's velocity when it strikes the ground is. D. 232.9 kg.m/s<span>.
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Read 2 more answers
A car of mass 1600 kg traveling at 27.0 m/s is at the foot of a hill that rises vertically 135 m after travelling a distance of
Zarrin [17]

Answer:

Neglecting any frictional losses, the average power delivered by the car's engine is 10565 W

Explanation:

The energy conservation law indicates that the energy must be the same at the bottom of the hill and at the top of the hill.  

The energy at the bottom is only the Kinect energy (K_1) of the car in motion, but in the top, the energy is the sum of its Kinect energy (K_2), potential energy (P) and the work (W) done by the engine.

K_1 = K_2 + P + W

then, the work done by the engine is:

W = K_1 - K_2 - P

The formulas for the Kinetic and potential energy are:  

K=\frac{1}{2}mV^2\\P=mgh

where, m is the mass of the car, V the velocity, g the gravity and h is the elevation of the hill.

Using the formulas:

W=\frac{1}{2}mV_1^2-\frac{1}{2}mV_2^2-mgh

Replacing the values:

W=\frac{1}{2}(1600Kg)(27m/s)^2-\frac{1}{2}(1600Kg)(14m/s)^2-(1600Kg)(9.8m/s^2)(135m)\\W=-1690400 J

The negative of this value indicates the direction of the work done, but for the problem, you only care about the magnitude, so the power is W=1690400 J. Now, the power is equal to work/time so you need to find the time the car took to get to the top of the hill.

The average speed of the car is (27+14)/2=20m/s, and t=d/v so the time is:

t=\frac{3200m}{20m/s}=160s

the power delivered by the car's engine was:

power=\frac{work}{time}=\frac{1690400J}{160s}=10565W

8 0
3 years ago
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