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bearhunter [10]
3 years ago
11

a 7 kg object moving 10 m/s Right collides with a 14 kg object at rest. If after the collision the 7kg object is at rest and the

14 kg object is moving, what is the velocity of the 14 kg object after the collision?
Physics
1 answer:
Lesechka [4]3 years ago
4 0

Answer:

v2(final)=5 m/s

Explanation:

we are going to use the conservation of momentum here

m1*v1(initial)+m2*v2(initial)=m1*v1(final)+m2v2(final)

m1=7 kg  v1(initial)=10 m/s

m2=14 kg  v2(initial)=0 m/s (bc initially it is at rest)

v1(final)= 0 m/s (m1 stops moving after the collision)

v2(final)=?

7*10+14*0=7*0+14*v2(final)

70=14v2(final)

v2(final)=70/14 m/s=5 m/s

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Answer:

Option 4

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Ratio of resistance of heater A to resistance of heater B is 5.80

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\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }

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\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }    ...(1)

The relation to determine electrical power for both heaters A and B are:

P_{1}=\frac{V^{2} }{R_{1} }     and

P_{2}=\frac{V^{2} }{R_{2} }

Here V is the voltage applied to both the heaters and is equal.

So, the ratio of electrical power of heaters is:

\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }     ....(2)

But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,

\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }

Substitute the suitable values in the above equation.

\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }

\frac{R_{1} }{R_{2} }=5.80

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