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.
Answer:
Explanation:
Given,
Width of slit, W = 5.7 x 10⁻⁴ m
Distance between central bright fringe, L = 4 m
distance between central bright fringe and first dark fringe, y = 4 mm
Diffraction angle
Now.
m = 1
The answer is D-Testable
Hope this helps
T= 3.34
Vi= 0
A= 9.81
D= ?
d=Vit+1/2at^2
d= 1/2(9.81)(3.34)2
d= 54.7 or 55 meters tall
Answer:
The current is 
Explanation:
From the question we are told that
The length of the segment is 
The current is 
The force felt is 
The distance of the second wire is 
Generally the current on the second wire is mathematically represented as
Here 
is the permeability of free space with value 
=> 
=>
