Answer:
Explanation:
Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.
1 ) Common acceleration a = force / total mass
= 234 / ( 25 +91 )
= 2.017 m s⁻².
2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂
= mass x acceleration
= 25 x 2.017
= 50.425 N
The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.
3 ) Maximum friction force that is possible between m₁ and m₂
= μ_s m₁g
= .79 x 25 x 9.8
= 193.55 N
Acceleration of m₁
= 193 .55 / 25
= 7.742 m s⁻²
This is the common acceleration in case of maximum tension required
So tension in rope
= ( 25 +91 ) x 7.742
= 898 N
4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁
= μ_k m₁g
= .62 x 25 x 9.8
= 151.9 N
Acceleration of m₁
= 151.9 / 25
= 6.076 m s⁻².
B is the correct option.
1. Given eqn;
S(t) = 1/2t² - 4t + 8
2.Differentiate the above eqn with respect to t;
<u>d(S(t))</u> = t - 4
dt
When distance, S, is differentiated it results to velocity.
V = t - 4
at t = 10
V = 10 - 4
V = 6 feet/s
Answer:
Option C is the correct answer
Explanation:
Distance travelled by car during reaction time
The car stopped before hitting the animal by 
Distance travelled during deceleration is 
Hence by 
We have
Option C is the correct answer
