The speed change : Δv = 0.41 m/s
<h3>Further explanation</h3>
Given
mass = 5.5 kg
Force = 15 N
time = 0.15 s
Required
the speed change
Solution
Newton 2nd's law
Impulse and momentum
F = m.a
F = m . Δv/t
F.t = m.Δv
Input the value :
15 N x 0.15 s = 5.5 kg x Δv
Δv = 0.41 m/s
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
Maurice directs a beam of light on a sheet of glass at an angle of 51°. The refractive index of glass is <span>1.46. </span>The angle of refraction in the glass is 29 degrees. The answer is letter B.
Answer:
77J
Explanation:
Not really an explanation to this, I just had this lesson last year and remembered it.
Hope I helped! ☺
