True, they represent the direction of motion
Answer:
The maximum displacement of the mass m₂ 
Explanation:
Kinetic Energy (K) = 1/2mv²
Potential Energy (P) = mgh
Law of Conservation of energy states that total energy of the system remains constant.
i.e; Total energy before collision = Total energy after collision
This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.
d = maximum displacement of the mass m₂
Answer:
50 m/s
Explanation:
Angle = 60 degree
Horizontal component of velocity = 50 m/s
A projectile motion is the motion of an object in two dimensions under the influence of gravity.
In this case, the object has no acceleration along horizontal direction, it has acceleration in vertical direction which is equal to the acceleration due to gravity of earth.
When the projectile reaches at the maximum height it travels only along the horizontal and thus it has only horizontal velocity at that instant.
Thus, the velocity of teh projectile at maximum height is same as horizontal component of velocity that meas 50 m/s.
Part A:
Acceleration can be calculated by dividing the difference of the initial and final velocities by the given time. That is,
a = (Vf - Vi) / t
where a is acceleration,
Vf is final velocity,
Vi is initial velocity, and
t is time
Substituting,
a = (9 m/s - 0 m/s) / 3 s = 3 m/s²
<em>ANSWER: 3 m/s²</em>
Part B:
From Newton's second law of motion, the net force is equal to the product of the mass and acceleration,
F = m x a
where F is force,
m is mass, and
a is acceleration
Substituting,
F = (80 kg) x (3 m/s²) = 240 kg m/s² = 240 N
<em>ANSWER: 240 N </em>
Part C:
The distance that the sprinter travel is calculated through the equation,
d = V₀t + 0.5at²
Substituting,
d = (0 m/s)(3 s) + 0.5(3 m/s²)(3 s)²
d = 13.5 m
<em>ANSWER: d = 13.5 m</em>
Answer:
the tangential velocity of the student is 4.89 m/s.
Explanation:
Given;
the radius of the circular path, r = 3.5 m
duration of the motion, t = 4.5 s
let the student's tangential velocity = v
The tangential velocity of the student is calculated as follows;
Therefore, the tangential velocity of the student is 4.89 m/s.
