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:
24mph
Explanation:
it really depends how high but the average speed for that quick will be atleast 24mph if not try 42mph if it is wrong
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Answer:
0.2932 rad/s
Explanation:
r = Radius = 2 m
= Initial angular momentum = 
= Initial angular velocity = 14 rev/min
= Final angular momentum
= Final angular velocity
Here the angular momentum of the system is conserved
The final angular velocity is 0.2932 rad/s
Answer:
Explanation:
This is a simple gravitational force problem using the equation:
where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.
Filling in:
I'm going to do the math on the top and then on the bottom and divide at the end.
and now when I divide I will express my answer to the correct number of sig dig's:
6.45 × 10¹⁶ N
