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Ymorist [56]
3 years ago
8

A large electrical device is plugged into a 240-volt outlet and has a resistance of 160 ohms. How much power does the device use

.
A.38,000 watts
B.1.5 watts
C.360 watts
D.100 watts
E.0.67 watts
Physics
2 answers:
Darya [45]3 years ago
6 0
The answer is simple 360 watts
lubasha [3.4K]3 years ago
5 0
Yur answer is c.360 watts
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A parachutist's falls to Earth is determined by two opposing forces. A gravitational force of 605 N acts on the parachutist. Aft
Juliette [100K]

Answer:

V_f=22.596\ m/sec

Explanation:

<u>Physics Dynamics </u>

The second Newton's Law, states the acceleration of a body will depend on the net force applied to it and its mass. If an object is left on free air the net force acting on it is the gravitational force. It will continue to fall with accelerated motion until that force is changed.

The formulas needed to compute the physics dynamics magnitudes are

F=ma

V_f=V_o+at

W=mg

The variables involved are: F=net force, m=mass, a=acceleration, V_f = final velocity, V_o = initial velocity, t = time, W = Weigh, g=9.8\ m/sec^2

At first, only the gravitational force of 605 N acts on the parachutist. That net force is due to the parachutist's weigh. We can know the mass

\displaystyle m=\frac{W}{g}=\frac{605N}{9.8m/sec^2}=61,735\ kg

If we assume the initial speed is 0, then

V_f=gt

All variables are assumed to be positive downwards. So, when t=3 sec

V_f=(9.8)(3)=29.4\ m/sec

Right then an air resistance force of 665 N appears. The new net force is

F=605N-665N=-60N

The new acceleration will be

\displaystyle m=\frac{F}{m}=\frac{-60N}{61,735\ kg}=-0.972\ m/sec^2

The acceleration is now negative since it goes upward. We are required to compute the speed after 10 sec (not clear if it's after this last event or it comes from the initial condition). We assume those 10 sec come from the very beginning of the jump, so t=7 sec

V_f=29.4\ m/s-0.972\ m/sec^2(7\ sec)=22.596\ m/sec

If it was t= 10\ sec

V_f=29.4\ m/s-0.972\ m/sec^2(10\ sec)=19.68\ m/sec

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Meaning of power in physics
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4 years ago
After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.
GarryVolchara [31]

Answer:

Imp = 11.666\,\frac{kg\cdot m}{s}

Explanation:

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v_{A} \approx 19.304\,\frac{m}{s}

After collision:

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v_{B} \approx 17.153\,\frac{m}{s}

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]

Imp = 11.666\,\frac{kg\cdot m}{s}

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