Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
The value is 
Explanation:
From the question we are told that
The focal length of the objective is 
The focal length of the eyepiece is 
The tube length is 
Generally the magnitude of the overall magnification is mathematically represented as
Where 
is the objective magnification which is mathematically represented as
=> 
=> 
is the eyepiece magnification which is mathematically evaluated as
So
Answer:
This link was diagram
Explanation:
https://doubtnut.app.link/FnsNC80Dccb
It’s c because it’s not Control so that means that it would be broken and non fix able
