Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
<em>Where</em>:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
<em>Where:</em>
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
Now, we can find the work:
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
50 strands is the standard procedure
Answer:
Torque on the rocket will be 1.11475 N -m
Explanation:
We have given that muscles generate a force of 45.5 N
So force F = 45.5 N
This force acts on the is acting on the effective lever arm of 2.45 cm
So length of the lever arm d = 2.45 cm = 0.0245 m
We have to find torque
We know that torque is given by 
So torque on the rocket will be 1.11475 N -m
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
