A spring with a mass of 5 Kg has natural length 0.5m. A force of 35.6 N is required to maintain it stretched to a length of 0.5m
1 answer:
Answer:
Explanation:
force constant of spring k = force / extension
= 35.6 / 0.5
k = 71.2 N / m
angular frequency ω of oscillation by spring mass system
where m is mass of the body attached with spring
Putting the values
ω = 3.77 radian / s
The oscillation of the mass will be like SHM having amplitude of 0.5 m and angular frequency of 3.77 radian /s . Initial phase will be π / 2
so the equation for displacement from equilibrium position that is middle point can be given as follows
x = .5 sin ( ω t + π / 2 )
= 0.5 cos ω t
= 0.5 cos 3.77 t .
x = 0.5 cos 3.77 t .
You might be interested in
Answer:
Energy= 46.08KJ
Explanation:
Given that the power needed to light each bulb is 32W
We know that Power =
The energy needed to light one bulb=
Given time = 1minute = 60 seconds
Energy = =1920J
Therefore energy needed to light one bulb is 1920J
The energy needed to light 24 bulbs = =46080J=46.08KJ
Answer:
Explanation:
Power is defined as the energy produced (E) per unit of time (t):
This means that the energy produced in the Sun each second (1 s), given the power , is
Each p-p chain reaction produces an amount of energy of
in order to get the total number of p-p chain reactions per second, we need to divide the total energy produced per second by the energy produced by each reaction:
The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude: