Answer:
<h2>1.17 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula

f is the force
m is the mass
From the question we have

We have the final answer as
<h3>1.17 m/s²</h3>
Hope this helps you
Decreases density because gases have less density than liquids
..............................................A
Let
denote the position vector of the ball hit by player A. Then this vector has components

where
is the magnitude of the acceleration due to gravity. Use the vertical component
to find the time at which ball A reaches the ground:

The horizontal position of the ball after 0.49 seconds is

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector
of the ball hit by player B has

Again, we solve for the time it takes the ball to reach the ground:

After this time, we expect a horizontal displacement of 12 meters, so that
satisfies


Explanation:
They probably put "rolls without slipping" in there to indicate that there is no loss in friction; or that the friction is constant throughout the movement of the disk. So it's more of a contingency part of the explanation of the problem.
(Remember how earlier on in Physics lessons, we see "ignore friction" written into problems; it just removes the "What about [ ]?" question for anyone who might ask.)
In this case, you can't ignore friction because the disk wouldn't roll without it.
As far as friction producing a torque... I would say that friction is a result of the torque in this case. And because the point of contact is, presumably, the ground, the friction is tangential to the disk. Meaning the friction is linear and has no angular component.
(You could probably argue that by Newton's 3rd Law there should be some opposing torque, but I think that's outside of the scope of this problem.)
Hopefully this helps clear up the misunderstanding for you.