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bija089 [108]
3 years ago
5

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

l moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Physics
1 answer:
Hoochie [10]3 years ago
5 0

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

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Answer:

1. 13..6 grams per centimeters cubed.

2. Mass = 400kg

3. Volume = 200cm = 2m

Explanation:

1. The conversion for kg/m^3 to g/cm^3 is divide by 1000.

2. density=\frac{mass}{volume}

1000=\frac{mass}{2 * 1 * 0.2}

1000*0.4=mass

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3. density=\frac{mass}{volume}

0.6=\frac{120}{volume}

volume=\frac{120}{0.6}

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In a certain time interval, natural gas with energy content of 19000 J was piped into a house during a winter day. In the same t
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Answer:

The amount of heat transfer is 21,000J .

Explanation:

The equation form of thermodynamics is,

ΔQ=ΔU+W

Here, ΔQ is the heat transferred, ΔU is the change in internal energy, and W is the work done.

Substitute 0 J for W and 0 J for ΔU

ΔQ = 0J+0J

ΔQ = 0J

The change in internal energy is equal to zero because the temperature changes of the house didn’t change. The work done is zero because the volume did not change

The heat transfer is,

ΔQ=Q  (in ) −Q (out )

Substitute 19000 J + 2000 J for Q(in)  and 0 J for Q(out)

ΔQ=(19000J+2000J)−(0J)

=21,000J

​Thus, the amount of heat transfer is 21,000J .

​

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