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IrinaVladis [17]
3 years ago
6

A strong man is compressing a lightweight spring between two weights. One weight has a mass of 2.3 kgkg , the other a mass of 5.

3 kgkg . He is holding the weights stationary, but then he loses his grip and the weights fly off in opposite directions. The lighter of the two is shot out at a speed of 6.0 m/sm/s .
Physics
1 answer:
Nezavi [6.7K]3 years ago
3 0

Incomplete question.The complete question is here

A strong man is compressing a lightweight spring between two weights. One weight has a mass of 2.3 kgkg , the other a mass of 5.3 kgkg . He is holding the weights stationary, but then he loses his grip and the weights fly off in opposite directions. The lighter of the two is shot out at a speed of 6.0 m/sm/s .What is the speed of the heavier weight?

Answer:

v_{2f}=2.60m/s

Explanation:

Given data

First weight mass m₁=2.3 kg

Second weight mass m₂=5.3 kg

Lighter weight speed v₁i= -6.0 m/s

To find

Heavier weight velocity v₂f

Solution

From law of conservation of momentum

p_{f}=p_{i}\\ m_{1}v_{1f}+m_{2}v_{2f}=m_{1}v_{1i}+m_{2}v_{2i}\\as\\v_{1i}=v_{2i}=0\\So\\ m_{1}v_{1f}+m_{2}v_{2f}=0\\ m_{1}v_{1f}=-m_{2}v_{2f}\\v_{2f}=(\frac{ m_{1}v_{1f}}{-m_{2}} )\\v_{2f}=(\frac{ 2.3kg*(-6.0m/s)}{-(5.3kg)} )\\v_{2f}=(-13.8/-5.3)\\v_{2f}=2.60m/s

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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
Dmitrij [34]

Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

(b) It must be placed inside a hollow shell.

Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

(iii) The direction of the net force is to the right i.e towards m_{2}.

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
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