Its been some time so i might be wrong but i think the answer is 3 either or 2
Answer:
65 N
Explanation:
Assuming that the ball was at rest hence initial momentum = 0
Final momentum = mv = 1Kg × 13 ms-1 = 13 kgms-1
But
Ft= mv according to Newton's law
F= mv/t
F= 13kgms-1/0.2s
F= 65 N
Answer:
A) the maximum acceleration the boulder can have and still get out of the quarry
B) how long does it take to be lifted out at maximum acceleration if it started from rest
Explanation:
A)
let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.
the weight of the chain is:
and maximum tension is ![T=2.50 m_{c} g=1.41*10^4N](https://tex.z-dn.net/?f=T%3D2.50%20m_%7Bc%7D%20g%3D1.41%2A10%5E4N)
total mass and weight is :
![M =m_{c}+ m_{b} =740kg+550kg=1290 kg](https://tex.z-dn.net/?f=M%20%3Dm_%7Bc%7D%2B%20m_%7Bb%7D%20%3D740kg%2B550kg%3D1290%20kg)
![w_{M} =1.2650*10^4N](https://tex.z-dn.net/?f=w_%7BM%7D%20%3D1.2650%2A10%5E4N)
∑![F_{y} =ma_{y}](https://tex.z-dn.net/?f=F_%7By%7D%20%3Dma_%7By%7D)
![T-M_{g} =Ma_{y}](https://tex.z-dn.net/?f=T-M_%7Bg%7D%20%3DMa_%7By%7D)
![a_{y} =(t-M_{g} )/M=(2.50m_{c} -M_{g} )/M=(2.50.550kg-1290kg)(9.8m/s^2)/1290kg](https://tex.z-dn.net/?f=a_%7By%7D%20%3D%28t-M_%7Bg%7D%20%29%2FM%3D%282.50m_%7Bc%7D%20-M_%7Bg%7D%20%29%2FM%3D%282.50.550kg-1290kg%29%289.8m%2Fs%5E2%29%2F1290kg)
![=0.645m/s^2](https://tex.z-dn.net/?f=%3D0.645m%2Fs%5E2)
B)
maximum acceleration
![a_{y} =0.645m/s^2\\\\y-y_{0} =119m\\v_{0y} =0](https://tex.z-dn.net/?f=a_%7By%7D%20%3D0.645m%2Fs%5E2%5C%5C%5C%5Cy-y_%7B0%7D%20%3D119m%5C%5Cv_%7B0y%7D%20%3D0)
using ![y-y_{0} =v_{oy} t+1/2(a_{y} )t^2](https://tex.z-dn.net/?f=y-y_%7B0%7D%20%3Dv_%7Boy%7D%20t%2B1%2F2%28a_%7By%7D%20%29t%5E2)
to solve for t
![t=\sqrt{2(y-y_{0} )/a_{y} }](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B2%28y-y_%7B0%7D%20%29%2Fa_%7By%7D%20%7D)
![t=\sqrt{2(119m)/0.645m/s^2} =19.20s](https://tex.z-dn.net/?f=t%3D%5Csqrt%7B2%28119m%29%2F0.645m%2Fs%5E2%7D%20%3D19.20s)