Answer: 5,640 s (94 minutes)
Explanation:
the tangential speed of the HST is given by
(1)
where
is the length of the orbit
r is the radius of the orbit
T is the orbital period
In our problem, we know the tangential speed:
. The radius of the orbit is the sum of the Earth's radius and the distance of the HST above Earth's surface:

So, we can re-arrange equation (1) to find the orbital period:

Dividing by 60, we get that this time corresponds to 94 minutes.
A I believe it would be true
Answer:

Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
![\displaystyle F_a=F_{r1}+F_{r2}.....[1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_a%3DF_%7Br1%7D%2BF_%7Br2%7D.....%5B1%5D)
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

The friction forces are computed by


Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]

Simplifying

Plugging in the values
![\displaystyle F_{a}=0.25(9.8)[400+2(100)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20F_%7Ba%7D%3D0.25%289.8%29%5B400%2B2%28100%29%5D)

Answer:

Explanation:
= Velocity of one lump = 
= Velocity of the other lump = 
m = Mass of each lump = 
The collision is perfectly inelastic as the lumps stick to each other so we have the relation

The velocity of the stuck-together lump just after the collision is
.