Answer:
5m/s2
Explanation:
Data obtained from the question include:
m (mass) = 2000 kg
F (force) = 10000N
a (acceleration) =?
Using the formula F = ma, the acceleration of the truck can be obtained as follow:
F = ma
a = F/m
a = 10000/2000
a = 5m/s2
The acceleration of the truck is 5m/s2
Answer:
Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F
Explanation:
We apply the concepts related to Power and energy stored in a capacitor.
By definition we know that power is represented as
Where,
E= Energy
t = time
to find the Energy we have,
With the energy found we can know calculate the Capacitance in a capacitor through the energy for capacitor equation, that is
Therefore the capacitance of the capacitor is 2.38 * 10⁻⁵F
Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).
Answer:
a) v = 5.88 m/s
b) v = 11.76 m/s
c) s = 1.764 m
d) s = 7.056 m
Explanation:
Given,
The initial velocity of the steel ball, u = o
The acceleration due to gravity, g = 9.8 m/s²
The equations of motion to find the final velocity for a body under free fall with an initial velocity, u = 0 is
v = u + at m/s
v = at m/s
a) At time t = 0.6 s
v = 9.8 x 0.6
= 5.88 m/s
The velocity of the ball 0.6 seconds after its release is, v = 5.88 m/s
b) At t = 1.2 s
v = 9.8 x 1.2
= 11.76 m/s
The velocity of the ball 1.2 seconds after its release is, v = 11.76 m/s
The distance traveled by the free falling body is given by the formula
s = ut + 1/2 at² m
s = 1/2 at² ∵ u = 0
c) At, t = 0.6 s
s = 1/2 x 9.8 x 0.6²
= 1.764 m
The distance ball fall in the first 0.6 seconds of its flight is, s = 1.764 m
d) At, t = 1.2 s
s = 1/2 x 9.8 x 1.2²
= 7.056 m
The distance ball fall in the first 1.2 seconds of its flight is, s = 7.056 m
