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inna [77]
2 years ago
12

A pole-vaulter converts the kinetic energy of running to elastic potential energy in the pole, which is then converted to gravit

ational potential energy. If a pole-vaulter's center of gravity is 1.3 m above the ground while he sprints at 10.3 m/s, what is the maximum height of his center of gravity during the vault? For an extended object, the gravitational potential energy is U = mgh, where h is the height of the center of gravity
Physics
1 answer:
ZanzabumX [31]2 years ago
6 0

Explanation:

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The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

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ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

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Which of the following objects exerts the greatest gravitational force on the Earth?
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Answer:

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A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
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Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

So final speed \frac{3.49\times 75}{100}=2.6175rad/sec

Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

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So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

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