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.
Sound waves travel around the boxed room causing them to bounce of the nearest walls to the end of the room>
Answer:
Before start of slide velocity will be 14.81 m/sec
Explanation:
We have given coefficient of static friction 
Angle of inclination is equal to 
Radius is given r = 28 m
Acceleration due to gravity 
We know that 
So before start of slide velocity will be 14.81 m/sec
<span>anwser will be
F = ma
where
F = force exerted on the bullet
m = mass of the bullet = 5 gm (given) = 0.005 kg.
a = acceleration of the bullet
Substituting appropriately,
F = 0.005a --- call this Equation 1
Next working equation is
Vf^2 - Vo^2 = 2as
where
Vf = velocity of the bullet as it leaves the muzzle = 326 m/sec (given)
Vo = initial velocity of bullet = 0
a = acceleration of bullet
s = length of the rifle's barrel
Substituting appropriately,
326^2 - 0 = 2(a)(0.83)
a = 64,022 m/sec^2
the anwser will be
Substituting this into Equation 1,
F = 0.005(64,022)
F =320.11 Newtons
Hope this helps. </span><span>
</span>
<span>Visible satellite images are like photos which are dependent on visible
light from the sun so they work best during the day. The sensor works by
detecting radiation within the range that wavelength is visible. Because of
this, the rays is usually seen as reaching earth from the East. </span>
