Answer:
Work done will be 2.205 j
Explanation:
We have given that the spring is compressed b 37.5 cm
So d = 0.375 m
Mass of the block m = 600 gram = 0.6 kg
Acceleration due to gravity 
Gravitational force on the block 
Now we know that work done is give by 
Answer:
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Explanation:
This is ur answer...
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Answer:
0.191 s
Explanation:
The distance from the center of the cube to the upper corner is r = d/√2.
When the cube is rotated an angle θ, the spring is stretched a distance of r sin θ. The new vertical distance from the center to the corner is r cos θ.
Sum of the torques:
∑τ = Iα
Fr cos θ = Iα
(k r sin θ) r cos θ = Iα
kr² sin θ cos θ = Iα
k (d²/2) sin θ cos θ = Iα
For a cube rotating about its center, I = ⅙ md².
k (d²/2) sin θ cos θ = ⅙ md² α
3k sin θ cos θ = mα
3/2 k sin(2θ) = mα
For small values of θ, sin θ ≈ θ.
3/2 k (2θ) = mα
α = (3k/m) θ
d²θ/dt² = (3k/m) θ
For this differential equation, the coefficient is the square of the angular frequency, ω².
ω² = 3k/m
ω = √(3k/m)
The period is:
T = 2π / ω
T = 2π √(m/(3k))
Given m = 2.50 kg and k = 900 N/m:
T = 2π √(2.50 kg / (3 × 900 N/m))
T = 0.191 s
The period is 0.191 seconds.
Answer:
If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
Explanation:
Option A is incorrect because, given this case, it is easier to calculate the field.
Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated
Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.
Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."
