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RUDIKE [14]
3 years ago
8

A student starts a food fight by throwing a 0.5 kg burrito at some girl he likes. He throws it kind of hard so he accelerates it

at 3 m/s^2 how much force did this take?
Physics
1 answer:
Maru [420]3 years ago
3 0

m = mass of burrito thrown by the student = 0.5 kg

a = acceleration of the burrito thrown by the student = 3 m/s²

F = force applied by the student on the burrito = ?

According to newton's second law , the net force on an object is the product of its mass and acceleration. it is given as

F = ma

inserting the values

F = (0.5) (3)

F = 1.5 N

hence the net force on the burrito comes out to be 1.5 N

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1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

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substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

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In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

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See Explanation

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The general explanation is as follows:

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