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nekit [7.7K]
3 years ago
9

A particular grocery bag can withstand 40 N of tension force before it rips. If a shopper places 3 kg of food in the bag, what i

s the maximum acceleration the bag can withstand before ripping?
Physics
2 answers:
snow_lady [41]3 years ago
8 0

Answer;

    = 13.33 m/s²

Solution;

From the second Newton's law of motion;

F= ma

F is the force = 40 N

M is the mass = 3 kg

a is the acceleration;

40 = 3 × a

a = 40/3

    = 13.33 m/s²

Hence; the maximum acceleration the bag can withstand before ripping is 13.33 m/s²

Lesechka [4]3 years ago
3 0

Answer:

3.3 is right

Explanation:

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m1 = 125 kg     v1 = 12m/s      v'1 = -12.5m/s
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m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'
(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'
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Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation. 

-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')
7.5kg.m/s=(235kg)(v_{2}')
(7.5kg.m/s)/(235kg)=(v_{2}')
0.03m/s=(v_{2}')

The velocity of the 2nd car after the collision is 0.03m/s.
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