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3 years ago

A 2,000 kg car is parked at the top of a 30 m high hill. what is its potential energy?

Physics

2 answers:

Elan Coil [88]3 years ago

8 0

Answer: The potential energy would be 600kiloJoule

Explanation: The Potential energy of a body is a function of it's wieght mg and height h. That is

Potential energy,

P.E = mass* acceleration due to gravity* height

Acceleration due to gravity is a constant 10m/s²

Potential energy = 2000*10*30 =600000Joule which is also 600kilojoule

Phoenix [80]3 years ago

6 0

The PE for this question will be 588,000 because we take the mass (2,000 kg), multiply it by 9.8 which is Gravitational Acceleration and then multiply that by the height (30 meters)

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The voltage V in a simple circuit is related to the current I and the resistance R by Ohm's Law: V=IR. An illustration of a simp

LenaWriter [7]

Answer:

dI/dt = 0.0004 A/s

Explanation:

R = 600 ohms

I = 0.04 A

dR/dt = -0.5 ohms/s

dV/dt = 0.04 V/s

From Ohm's law V = IR

Taking the derivative of both sides with respect to t using product rule

dV/dt = I dR/dt + R dI/dt

0.04 = -( 0.04*0.5) + (600) dI/dt

0.04 + 0.2 = 600 dI/dt

0.24 = 600 dI/dt

dI/dt = 0.24/600

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A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

djverab [1.8K]

Answer:

0.005 V

Explanation:

We are given that

Initial circumference of circular loop=C=165 cm

Rate of circumference, $\frac{dC}{dt}=12 cm/s$

Magnetic field,B=0.5 T

We have to find the induced emf at time t=9 s

We know that induced amf,E= $\frac{Bd(A)}{dt}$

Area of circular coil,A= $\pi r^2$

$E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}$

Circumference of circular coil,C= $2\pi r$

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$r=\frac{165}{2\pi}$

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$r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m$

$1 m=100 cm$

E= $-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V$

Magnitude of induced emf=0.005 V

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