Answer:
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Applying the impulse-momentum equation;
Impulse = change in momentum
Ft = m∆v
F = (m∆v)/t
Where;
F = force
t = time
m = mass
∆v = v2 - v1 = change in velocity
Given;
m = 0.80 kg
t = 0.050 s
The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.
v2 = 25 m/s
v1 = -25 m/s
∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s
Substituting the values;
F = (m∆v)/t
F = (0.80×50)/0.05
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
kinematic equation
v=u+at
v-u=at
v-u = 1x5
the driver will have increased speed by 5 m/s. actual speeds unknown
The linear scale is applicable only as it moves in one dimension. From the word "linear" it means it deals with one equation only. Unlike the other options, the dimensions are many because it involves 2 or more variables for its equation.
A, something to remember is that if the numbers are even the answer too will be even. Every time hope this helps :>
Please ignore my comment -- mass is not needed, here is how to solve it. pls do the math
at bottom box has only kinetic energy
ke = (1/2)mv^2
v = initial velocity
moving up until rest work done = Fs
F = kinetic fiction force = uN = umg x cos(a)
s = distance travel = h/sin(a)
h = height at top
a = slope angle
u = kinetic fiction
work = Fs = umgh x cot(a)
ke = work (use all ke to do work)
(1/2)mv^2 = umgh x cot(a)
u = (1/2)v^2 x tan (a) / gh
