Answer:
28√3 m
Explanation:
A = vertex where receiver is placed
S = focus
Bp = r = radius of the outside edge
Bc = 2r = diameter
The full explanation is shown in the picture attached herewith. Thank you and i hope it helps.
The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
So by applying the theorem of parallel axes, the moment of inertia of the sphere is computed by:I = 2/5*mR^2 for rotation about the center of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere.
I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
Since I'm assuming that its perfectly elastic, considering there's not enough information given, so I think that no energy is dissipated in the collision
hmax = h - d + { [ mpvp - mb√(2gd) ] / (mp+mb) }² / (2g)
Answer:
The frequency of radiation is 
Explanation:
Given:
Wavelength 
m
Speed of light 
For finding the frequency of radiation,
Therefore, the frequency of radiation is
The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0 → <em>n</em> = <em>mg</em> = 1876 N
where <em>n</em> is the magnitude of the normal force.
In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,
∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0 → <em>f</em> = 747 N
The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that
<em>f</em> = <em>µn</em> → <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40
