The car’s velocity at the end of this distance is <em>18.17 m/s.</em>
Given the following data:
- Initial velocity, U = 22 m/s
- Deceleration, d = 1.4

To find the car’s velocity at the end of this distance, we would use the third equation of motion;
Mathematically, the third equation of motion is calculated by using the formula;

Substituting the values into the formula, we have;

<em>Final velocity, V = 18.17 m/s</em>
Therefore, the car’s velocity at the end of this distance is <em>18.17 m/s.</em>
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Read more: brainly.com/question/8898885
Answer:
42.6 m
Explanation:
mass of crate m = 53 kg
coefficient of kinetic friction, μ = 0.36
acceleration due to gravity, g = 9.8 m/s^2
Force, F = 372.098 N
Net force, f = F - friction force
f = 372.098 - μ m x g = 372.098 - 0.36 x 53 x 9.8
f = 185.114 N
acceleration, a = f / m = 185.114 / 53 = 3.49 m/s^2
initial velocity, u = 0
time, t = 4.94 s
s = ut + 1/2 at^2
s = 0 + 1/2 x 3.49 x 4.94 x 4.94
s = 42.6 m
Answer:
388.5J
Explanation:
Given parameters:
Weight = 70N
Height = 5.55m
Unknown:
Gravitational potential energy at the top of the ladder = ?
Solution:
The gravitational potential energy is the energy due to the position of the body.
Gravitational potential energy = Weight x height
So;
Gravitational potential energy = 70 x 5.55 = 388.5J
Answer:
A) 89.39 J
B) 30.39J
C) 23.8 J
Explanation:
We are given;
F = 30.2N
m = 3.5 kg
μ_k = 0.646
d = 2.96m
ΔEth (Block) = 35.2J
A) Work done by the applied force on the block-floor system is given as;
W = F•d
Thus, W = 30.2 x 2.96 = 89.39 J
B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;
ΔEth = μ_k•mgd
Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J
Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.
Thus,
ΔEth = ΔEth (Block) + ΔEth (floor)
Thus,
ΔEth (floor) = ΔEth - ΔEth (Block)
ΔEth (floor) = 65.59J - 35.2J = 30.39J
C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;
W = K + ΔEth
Therefore;
K = W - ΔEth
K = 89.39 - 65.59 = 23.8J
Answer:

Explanation:
The period of the simple pendulum is:

Where:
- Cord length, in m.
- Gravity constant, in
.
Given that the same pendulum is test on each planet, the following relation is formed:

The ratio of the gravitational constant on planet CornTeen to the gravitational constant on planet Earth is:


