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Answer:
The speed of the car at the end of the 2nd second = 8.0 m/s
Explanation:
The equations of motion will be used to solve this problem.
A car starts from rest,
u = initial velocity of the car = 0 m/s
Accelerates at a constant rate in a straight line,
a = constant acceleration of the car = ?
In the first second the car moves a distance of 2.0 meters,
t = 1.0 s
x = distance covered = 2.0 m
x = ut + (1/2)at²
2 = 0 + (1/2)(a)(1²)
a = 4.0 m/s²
How fast will the car be moving at the end of the second second
Now,
a = 4.0 m/s²
u = initial velocity of the car at 0 seconds = 0 m/s
v = final velocity of the car at the end of the 2nd second = ?
t = 2.0 s
v = u + at
v = 0 + (4×2)
v = 8.0 m/s
Rolling without slipping means V=rw where w is angular velocity. So 4.0=0.1w
w=40rad/s
Answer:
Final velocity of ball = 20 m/s
Explanation:
Given:
Height when ball released = 20meter
Find;
Time taken to fall
Final velocity of ball
Computation:
Initial velocity of ball = 0 m/s
Acceleration due to gravity = 10 m/s²
Using equation of motion;
S = ut + (1/2)(g)(t²)
20 = (0)(t) + (1/2)(10)(t²)
20 = 5(t²)
(t²) = 20 / 5
(t²) = 4
Time taken to fall = 2 second
V = u + at
V = 0 + (10)(2)
Final velocity of ball = 20 m/s
Answer:
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Explanation: