Answer:
D
Explanation:
I just had this question the answer is D
Answer:

Explanation:
From the law of conservation of energy
Energy lost by the spring, W=Kinetic energy gained, KE+Potential energy gained, PE+Work done by friction, Fr



The required distance from A to B is 
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
Thank you for posting your Physics question here. I hope the answer helps. Upon calculating the ramp with the horizontal the answer is 20.49 Deg. Below is the solution:
Y = 7 m.
<span>r = 20 m. </span>
<span>sinA = Y/r = 7/20 = 0.35. </span>
<span>A = 20.49 Deg.</span>
Answer:
The equation for the object's displacement is 
Explanation:
Given:
m = 16 lb
δ = 3 in
The stiffness is:

The angular speed is:

The damping force is:

Where
FD = 20 lb
u = 4 ft/s = 48 in/s
Replacing:

The critical damping is equal:

Like cc>c the system is undamped
The equilibrium expression is:
