Answer:
Fc = 19.2 N
Explanation:
In this case, the force of the Honda over the rock, is a centripetal force. Then, you have:
m: mass of the rock = 600g = 0.6 kg
v: tangential velocity of the Honda = 4m/s
r: radius of the Honda = 50cm = 0.5m
You replace the values of m, r and v in the equation for Fc:
hence, the force has a magnitude of 19.2 N
If the rock would have more mass the centripetal force would be higher
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
Answer:
R = 73.25 m
Explanation:
We have,
Initial speed of the ball is 27 m/s
It is projected at an angle of 40 degrees
The maximum range of the ball is given by :
Plugging all the values we get :
So, the maximum range of the ball is 73.25 m
Force = mass × acceleration = kg × m/s^2 = Newton