Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
Answer:
F = 768 N
Explanation:
It is given that,
Speed of the elevator, v = 3.2 m/s
Grain drops into the car at the rate of 240 kg/min, 
We need to find the magnitude of force needed to keep the car moving constant speed. The relation between the momentum and the force is given by :
Since, the speed is constant,
F = 768 N
So, the magnitude of force need to keep the car is 768 N. Hence, this is the required solution.
She can put chalk in vinegar as the vinegar will disintegrate the chalk chemically demonstrating chemical changes. But for physical changes she can break the chalk into small pieces by smashing it with something or her hand.
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A) No, the equations presented above are the product of the derivation of position and velocity when the acceleration is constant.
The equations change to polynomial function of the second degree for the description of the acceleration when described as a function of time.
B) Yes, when the acceleration is zero it is concluded that the velocity is constant, therefore they could be used to describe the position as a function of the change in velocity.
