Answer:
<h2>

</h2>
Explanation:


?

Now, let's find the mass:

plug the values
⇒
Multiply the numbers
⇒
Swap the sides of the equation
⇒
Divide both sides of the equation by 270.48
⇒
Calculate
⇒
kg
Hope I helped!
Best regards!!
Answer: 4 s
Explanation:
Given
The applied force is 70 N
mass of the rock is 28 kg
initial velocity 
final velocity 
Deceleration provided by force is

using the equation of motion

Answer:
Centrifugal force
Explanation:
The reacting force that is equal to and opposite in the direction to the centripetal force and tends to fling air out of the center of rotation of High and Low pressure systems is called centrifugal force.
Centrifugal force is force that causes an object moving in a circular path to move out and away from the center of it's path, it always centripetal force and the force is imaginary, which can only be felt and not seen.
Answer:
v= 13 m/s
Explanation:
Velocity is defined as the derivative of displacement with respect to time
v= ds/dt
Known data
s(t) = 5t + 2t² : distance that the ball has rolled after t seconds
vi= 5 m/s : initial velocity
t= 2 s
Problem develoment
s(t) = 5t + 2t²
v= ds/dt= 5 + 4t : velocity of the ball in function of the time
We replace t =2 s in the equation of velocity
v= 5 + 4(2)
v= 13 m/s : velocity after 2 seconds
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