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Marta_Voda [28]
3 years ago
6

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you

at 13.0 m/s. The maximum force you apply in catching the water balloon is twice the average force. How long must the interaction time of your catch be to make sure the water balloon doesn't soak you
Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}       (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s

The interaction time to avoid that the water balloon breaks is 0.029s

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For example, suppose you lift a suitcase of mass m through a height h. The weight W of the suit case is a downward force of size mg. In lifting the suitcase, you would have to pull upwards on it with a force equal in size to its weight, mg.

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