Answer:
4.18
Explanation:
Givens
The car's initial velocity 
= 0 and covering a distance Δx = 1/4 mi = 402.336 m in a time interval t = 4.43 s.
Knowns
We know that the maximum static friction force is given by:
μ_s*n (1)
Where μ_s is the coefficient of static friction and n is the normal force.
Calculations
(a) First, we calculate the acceleration needed to achieve this goal by substituting the given values into a proper kinematic equation as follows:
Δx=
a=41 m/s
This is the acceleration provided by the engine. Applying Newton's second law on the car, so in equilibrium, when the car is about to move, we find that:
Substituting (3) into (1), we get:
μ_s*m*g
Equating this equation with (4), we get:
ma= μ_s*m*g
μ_s=a/g
=4.18
<span>A cumulus cloud is puffy and white.
</span><span>Vinegar has a very sour smell.
</span><span>Water boils at 100 degrees Celsius. </span>
Answer:
The final velocity of the vehicle is 10.39 m/s.
Explanation:
Given;
acceleration of the vehicle, a = 2.7 m/s²
distance moved by the vehicle, d = 20 m
The final velocity of the vehicle is calculated using the following kinematic equation;
v² = u² + 2ah
v² = 0 + 2 x 2.7 x 20
v² = 108
v = √108
v = 10.39 m/s
Therefore, the final velocity of the vehicle is 10.39 m/s.
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s
