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: im not sire
Explanation: very sorry im not sure
<u>Answer:</u>
total mass = 410 g
<u>Explanation:</u>
density = 1.8 g/cm³
volume = 200 cm³
density = mass / volume
mass (of liquid) = density x volume
= 1.8 x 200
= 360 g
total mass (beaker + liquid) = 50 + 360 = 410 g [Ans]
Hope this helps!
The correct answer to the question is vertically downward i.e towards the centre of earth.
EXPLANATION:
As per the question, the box is pulled to the right.
Hence, the direction of the applied force is towards right.
We are asked to determine the direction of the gravitational force that acts on the body.
Before answering this question, first we gave to understand the gravitational force of earth.
Any body present on the surface of earth is attracted with the force of gravity of earth ( gravitational force ) towards its centre. It is equivalent to the weight of the body.
The force of gravity is always directed towards the centre of earth irrespective of the nature of applied force.
Hence, the direction of the gravitational force which acts on the box is vertically downward.
