Answer:
4.25 m/s
Explanation:
Force, F = 22 N
Time, t = 0.029 s
mass, m = 0.15 kg
initial velocity of the cue ball, u = 0
Let v be the final velocity of the cue ball.
Use newton's second law
Force = rate of change on momentum
F = m (v - u) / t
22 = 0.15 ( v - 0) / 0.029
v = 4.25 m/s
Thus, the velocity of cue ball after being struck is 4.25 m/s.
Answer:
101397.16 pa
Explanation:
The pressure recorded will be equal to pgh
Where p = density of mercury = 13.6x10^3 kg/m^ 3
g = acceleration due to gravity 9.81 m/s^2
h = height of mercury in the column = 760 mm = 760x10^-3 m
Pressure = 13.6x10^3 x 9.81 x 760x10^-3 = 101397.16 pa
Answer:

Explanation:
The x- and y- components of the velocity vector can be written as following:


Since the angle θ and the magnitude of the velocity is given, the vector representation can be written as follows:

In order to make the athletic shoe to slide faster of a given surface you had to reduce the friction coefficient of the shoe with the surface.
One way to do that is to put some grease of the sole of the shoe, which undoubtedly would reduce the friction coefficient drastically and the shoe would slide more easily.
Also, if you make the sole smoother the shoe the friction coefficient would decrease and, again, the show would slide mor easily.