Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
![a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7BF%7D%7Bm%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B5%7D%7B2%7D%5C%5C%5C%5Ca%3D2.5%5C%20m%2Fs%5E2)
Let d is the distance moved in 2.25 s. Using second equation of motion,
![d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m](https://tex.z-dn.net/?f=d%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5C%5Cd%3D0%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%202.5%5Ctimes%20%282.25%29%5E2%5C%5C%5C%5Cd%3D6.32%5C%20m)
So, it will move 6.32 m from rest in 2.25 seconds.
The best illustration that represents the interaction is D
Answer:
<h3>The answer is 8.91 m/s²</h3>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
![a = \frac{f}{m} \\](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7Bf%7D%7Bm%7D%20%20%5C%5C%20)
f is the force
m is the mass
From the question we have
![a = \frac{343}{38.5} = \frac{98}{11} \\ = 8.909090...](https://tex.z-dn.net/?f=a%20%3D%20%20%5Cfrac%7B343%7D%7B38.5%7D%20%20%3D%20%20%5Cfrac%7B98%7D%7B11%7D%20%20%5C%5C%20%20%3D%208.909090...)
We have the final answer as
<h3>8.91 m/s²</h3>
Hope this helps you