Answer:
1.4 m
Explanation:
v = Speed of sound in water = 1400 m/s
f = Frequency of sound = 1000 Hz
= Wavelength
When we multiply the frequency and the wavelength of a wave we get the velocity of sound in that medium
The wavelength of the sound waves in water is 1.4 m
<h2>
After 26.28 seconds projectile returns 26.28 seconds.</h2>
Explanation:
Initial velocity = 450 ft/s = 137.16 m/s
Angle, θ = 70°
Consider the vertical motion of projectile,
When the projectile return to the ground we have
Displacement, s = 0 m
Acceleration, a = -9.81 m/s²
Initial velocity, u = 137.16 x sin70 = 128.89 m/s
Substituting in s = ut + 0.5 at²
s = ut + 0.5 at²
0 = 128.89 x t + 0.5 x (-9.81) x t²
t² - 26.28 t = 0
t ( t- 26.28) = 0
t = 0 s or t = 26.28 s
After 26.28 seconds projectile returns 26.28 seconds.
Answer:
5.4 ms⁻¹
Explanation:
Here we have to use conservation of energy. Initially when the stick is held vertical, its center of mass is at some height above the ground, hence the stick has some gravitational potential energy. As the stick is allowed to fall, its rotates about one. gravitational potential energy of the stick gets converted into rotational kinetic energy.
= length of the meter stick = 1 m
= mass of the meter stick
= angular speed of the meter stick as it hits the floor
= speed of the other end of the stick
we know that, linear speed and angular speed are related as
= height of center of mass of meter stick above the floor = 
= Moment of inertia of the stick about one end
For a stick, momentof inertia about one end has the formula as
Using conservation of energy
Rotational kinetic energy of the stick = gravitational potential energy
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.
