Answer:
90 N
Explanation:
The force applied to the ball is given by:
![F=\frac{\Delta p}{\Delta t}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%5CDelta%20p%7D%7B%5CDelta%20t%7D)
where
is the change in momentum of the ball
is the time taken
The change on momentum of the ball is:
![\Delta p=m\Delta v=(0.45 kg)(20 m/s)=9 kg m/s](https://tex.z-dn.net/?f=%5CDelta%20p%3Dm%5CDelta%20v%3D%280.45%20kg%29%2820%20m%2Fs%29%3D9%20kg%20m%2Fs)
So, the force applied is
![F=\frac{9 kg m/s}{0.10 s}=90 N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B9%20kg%20m%2Fs%7D%7B0.10%20s%7D%3D90%20N)
Answer:
Explanation:
It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.
Let in equilibrium , tension in rope be T
For balancing Joe
T = M g
For balancing Simon
friction + T = mgsinθ
μmgcosθ+T = mgsinθ
μmgcosθ+Mg = mgsinθ
M = (msinθ - μmcosθ)
M = m(sinθ - μcosθ)
Answer:
During the section CD , the speed is fastest.
Explanation:
The rate of change of distance is called speed.
Speed = distance / time
Its SI unit ism/s. It is a scalar quantity.
The slope of the distance time graph is given by the speed of the object.
Here, the speed of AB is 30/3= 10 m/s .
The speed of BC is = 0 m/s
The speed of CD is (50 - 30)/(6 - 5) = 20 m/s
So, the speed is maximum during the section CD.
<span>A(n) alpha particle is a particle that contains two protons, two neutrons, and has a 2 charge.
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b. alpha