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From the information given, The mass of the bowling ball is 8 Kilograms and the momentum with which it is moving is 16 kg m/s.
We use the formula p = m × v
Where p is the momentum, m is the mass and v is the velocity.
We need velocity so we rewrite the equation thus:
P = mv, therefore p/m = v or v = p/m
In our case p = 16 and m = 8
v = p/m
v = 16/8
v = 2
Therefore the bowling ball is travelling at 2m/s
Answer:
Woke done, W = 4156.92 Joules
Explanation:
The work done by the force can be calculated as :
is the angle between force and the displacement
It is assumed to find the work done for the given parameters i.e.
Force, F = 30 N
Distance travelled, s = 160 m
Angle between force and displacement, 
Work done is given by :
W = 4156.92 Joules
So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
