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yan [13]
3 years ago
11

A mover hoists a 200 kg piano from the ground to a height of 10 m using a single pulley. What was the change in the pianos energ

y? a. 20j b. 200j c. 2000j d. 20000
Physics
2 answers:
RideAnS [48]3 years ago
7 0
P.E= Mass x gravity x height   (standard gravity = 10m/s approaximately)
     = 200x10x10= 20000j (D)
aivan3 [116]3 years ago
3 0

Answer: d. 20000 J

Explanation:

Potential energy is the energy possessed by an object by virtue of its position.

P.E=m\times g\times h

m= mass of object = 200 kg

g = acceleration due to gravity = 10ms^{-2}

h = height of an object

1. when h= 10 m

P.E=200\times 10ms^{-2}\times 10m=20000Joules

2. when h= 0 m

P.E=200\times 10ms^{-2}\times 0m=0Joules

Thus change in energy=(20000-0)Joules=20000Joules.

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An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linea
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Answer:

Velocity of the electron at the centre of the ring, v=1.37\times10^7\ \rm m/s

Explanation:

<u>Given:</u>

  • Linear charge density of the ring=0.1\ \rm \mu C/m
  • Radius of the ring R=0.2 m
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Total charge of the ring

Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C

Potential due the ring at a distance x from the centre of the rings is given by

V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V

Let\Delta U be the change in potential Energy given by

\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s

So the electron will be moving with v=1.37\times10^7\ \rm m/s

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3 years ago
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<span>The work output of a machine divided by the work input is the "Efficiency" of the machine.

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