a particle of mass2.5kg moves in a cnservative force field.its potential energy curveis dhwn below.from the curve,detrmine(a)the
1 answer:
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Explanation:
Given parameters:
Distance = 15miles north = 24140.2m
Initial velocity = 0m/s
Final velocity = 4m/s
Unknown:
Speed, velocity and acceleration = ?
Solution:
The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.
Speed =
The speed of the student is 4m/s
Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;
Velocity =
The velocity of the student is 4m/s due north
Acceleration is the change in velocity with time;
To find the acceleration, we use
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
4² = 0² + 2x a x 24140.2
a = = 0.00033m/s²
Answer:
The velocity of the ball when its hit the ground will be 54.22 m/sec
Explanation:
We have given height from which ball is dropped h = 150 m
Acceleration due to gravity
As the ball is dropped so initial velocity will be zero so u = 0 m/sec
According to third equation of motion we know that
So the velocity of the ball when its hit the ground will be 54.22 m/sec