here we will use the concept of Newton's III law
as per Newton's III law the impulse given to the ball is same as the impulse lost by the bat
So here we will say
impulse gain by the ball = impulse lost by the bat
given that
For ball the change in speed will be
now from above equation
so speed of bat will decrease by 6.72 mph
This electric force calculator will enable you to determine the repulsive or attractive force between two static charged particles. Continue reading to get a better understanding of Coulomb's law, the conditions of its validity, and the physical interpretation of the obtained result.
How to use Coulomb's law
Coulomb's law, otherwise known as Coulomb's inverse-square law, describes the electrostatic force acting between two charges. The force acts along the shortest line that joins the charges. It is repulsive if both charges have the same sign and attractive if they have opposite signs.
Coulomb's law is formulated as follows:
F = keq₁q₂/r²
where:
F is the electrostatic force between charges (in Newtons),
q₁ is the magnitude of the first charge (in Coulombs),
q₂ is the magnitude of the second charge (in Coulombs),
r is the shortest distance between the charges (in m),
ke is the Coulomb's constant. It is equal to 8.98755 × 10⁹ N·m²/C². This value is already embedded in the calculator - you don't have to remember it :)
Simply input any three values
An object in motion eventually stops because of friction.
<h3>What is Newton’s first law of motion?</h3>
Newton's first law states that if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or continue its moving in a straight line at constant speed unless force is applied on it.
So we can conclude that An object in motion eventually stops is the correct answer.
Learn more about law here: brainly.com/question/820417
Answer:
A.) 1430 metres
B.) 80 seconds
Explanation:
Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:
U = acceleration × time
U = 1.1 × 20 = 22 m/s
It then proceeds at constant speed for 1100 m
Then, time t will be
Time = distance/ velocity
Time = 1100/22
Time = 50 s
before slowing down at 2.2m/s^2 until it stops at the station.
Deceleration = velocity/time
2.2 = 22/t
t = 22/2.2
t = 10s
Using area under the graph, the distance between the two stations will be :
(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)
220 + 1100 + 110
1430 m
The time taken between the two stations will be
20 + 50 + 10 = 80 seconds
