Answer: I think the answer is D
Explanation: N/A
Answer:
4500 N
Explanation:
When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:
a = v^2/r
where v is the velocity of the body and r is the radius of the circumference:
Therefore, a body with mass m, will feel a force f:
f = m v^2/r
Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:
fs = μN = μmg
The car will not slide if f = fs, i.e.
fs = μmg = m v^2/r
That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration
fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N
I can guarantee you that it is not
C.<span>the angle that the incident ray makes with a line drawn perpendicular to the reflecting surface I hope this somewhat helps</span>
Answer:
The first part can be solved via conservation of energy.
For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.
where 
because we are looking for the case where the car loses contact.
Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.
Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.
Answer:
Vb = k Q / r r <R
Vb = k q / R³ (R² - r²) r >R
Explanation:
The electic potential is defined by
ΔV = - ∫ E .ds
We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product
VB - VA = - ∫ E dr
Let's substitute every equation they give us and we find out
r> R
Va = - ∫ (k Q / r²) dr
-Va = - k Q (- 1 / r)
We evaluate with it Va = 0 for r = infinity
Vb = k Q / r r <R
We perform the calculation of the power with the expression of the electric field that they give us
Vb = - int (kQ / R3 r) dr
We integrate and evaluate from the starting point r = R to the final point r <R
Vb = ∫kq / R³ r dr
Vb = k q / R³ (R² - r²)
This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity
