Answer:
D
Explanation:
The decrease in potential energy is equal to the increase in kinetic energy.
mgh
250 x 9.8 x 30
=73, 500
Answer:
The time it takes the ball to fall 3.8 meters to friend below is approximately 0.88 seconds
Explanation:
The height from which the student tosses the ball to a friend, h = 3.8 meters above the friend
The direction in which the student tosses the ball = The horizontal direction
Given that the ball is tossed in the horizontal direction, and not the vertical direction, the initial vertical component of the velocity of the ball = 0
The equation of the vertical motion of the ball can therefore, be represented by the free fall equation as follows;
h = 1/2 × g × t²
Where;
g = The acceleration due gravity of the ball = 9.81 m/s²
t = The time of motion to cover height, h
Then height is already given as h = 3.8 m
Substituting gives;
3.8 = 1/2 × 9.81 × t²
t² = 3.8/(1/2 × 9.81) ≈ 0.775 s²
∴ t = √0.775 ≈ 0.88 seconds
The time it takes the ball to fall 3.8 meters to friend below is t ≈ 0.88 seconds.
Answer:
Explanation:
Given that
d= 1.5 in ( 1 in = 0.0254 m)
d= 0.0381 m
P= 75 hp ( 1 hp = 745.7 W)
P= 55927.5 W
N= 1800 rpm
We know that power P is given as
T=Torque
N=Speed
T=296.85 N.m
The maximum shear stress is given as
We know that 1 MPa =0.145 ksi
Answer:
Final speed of the car, v = 24.49 m/s
Explanation:
It is given that,
Initial velocity of the car, u = 0
Acceleration, 
Time taken, t = 7.9 s
We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :
v = 24.49 m/s
So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.
Question:
A 63.0 kg sprinter starts a race with an acceleration of 4.20m/s square. What is the net external force on him? If the sprinter from the previous problem accelerates at that rate for 20m, and then maintains that velocity for the remainder for the 100-m dash, what will be his time for the race?
Answer:
Time for the race will be t = 9.26 s
Explanation:
Given data:
As the sprinter starts the race so initial velocity = v₁ = 0
Distance = s₁ = 20 m
Acceleration = a = 4.20 ms⁻²
Distance = s₂ = 100 m
We first need to find the final velocity (v₂) of sprinter at the end of the first 20 meters.
Using 3rd equation of motion
(v₂)² - (v₁)² = 2as₁ = 2(4.2)(20)
v₂ = 12.96 ms⁻¹
Time for 20 m distance = t₁ = (v₂ - v ₁)/a
t₁ = 12.96/4.2 = 3.09 s
He ran the rest of the race at this velocity (12.96 m/s). Since has had already covered 20 meters, he has to cover 80 meters more to complete the 100 meter dash. So the time required to cover the 80 meters will be
Time for 100 m distance = t₂ = s₂/v₂
t₂ = 80/12.96 = 6.17 s
Total time = T = t₁ + t₂ = 3.09 + 6.17 = 9.26 s
T = 9.26 s
