Answer:
Part a)

Part b)
v = 3.64 m/s
Part c)

Part d)

Explanation:
As we know that moment of inertia of hollow sphere is given as

here we know that

R = 0.200 m
now we have


now we know that total Kinetic energy is given as





Part a)
Now initial rotational kinetic energy is given as



Part b)
speed of the sphere is given as
v = 3.64 m/s
Part c)
By energy conservation of the rolling sphere we can say




Part d)
Now we know that




Density is mass divided by volume. rho=m/v. So, v=m/rho. In frank's case this is 80/8 = 10 cm^3.
Answer:
the first one is Primary
the second one I think it's Mature but I don't know
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m