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:
343/1500
Explanation:
Power: This can be defined as the product force and velocity. The S.I unit of power is Watt (w).
From the question,
P' = mg×v................. Equation 1
Where P' = power used to gain an altitude, m = mass of the engine, g = acceleration due to gravity of the engine, v = velocity of the engine.
Given: m = 700 kg, v = 2.5 m/s, g = 9.8 m/s²
Substitute into equation 1
P' = 700(2.5)(9.8)
P' = 17150 W.
If the full power generated by the engine = 75000 W
The fraction of the engine power used to make the climb = 17150/75000
= 343/1500
Answer:
Explanation:
Potential difference between two points in constant electric field is given by the formula
here we know that
also we know that
now we have
now change in potential energy is given as
Yes. An object of large mass is pulled down onto a surface with a greater force than an object of low mass and, as a consequence, there is greater friction between the surface of the heavy object than between the surface of the light object.
