The moment of inertia of a point mass about an arbitrary point is given by:
I = mr²
I is the moment of inertia
m is the mass
r is the distance between the arbitrary point and the point mass
The center of mass of the system is located halfway between the 2 inner masses, therefore two masses lie ℓ/2 away from the center and the outer two masses lie 3ℓ/2 away from the center.
The total moment of inertia of the system is the sum of the moments of each mass, i.e.
I = ∑mr²
The moment of inertia of each of the two inner masses is
I = m(ℓ/2)² = mℓ²/4
The moment of inertia of each of the two outer masses is
I = m(3ℓ/2)² = 9mℓ²/4
The total moment of inertia of the system is
I = 2[mℓ²/4]+2[9mℓ²/4]
I = mℓ²/2+9mℓ²/2
I = 10mℓ²/2
I = 5mℓ²
If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.
From the calculations, the power expended is 43650 W.
<h3>What is the power expended?</h3>
Now we can find the acceleration from;
v = u + at
u = 0 m/s
v = 95 km/h or 26.4 m/s
t = 6.8 s
a = ?
Now
v = at
a = v/t
a = 26.4 m/s/ 6.8 s
a = 3.88 m/s^2
Force = ma = 850-kg * 3.88 m/s^2 = 3298 N
The distance covered is obtained from;
v^2 = u^2 + 2as
v^2 = 2as
s = v^2/2a
s = (26.4)^2/2 * 3.88
s = 696.96/7.76
s = 90 m
Now;
Work = Fs
Work = 3298 N * 90 m = 296820 J
Power = 296820 J/ 6.8 s
= 43650 W
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Answer:
the velocity of the boats after the collision is 4.36 m/s.
Explanation:
Given;
mass of fish, m₁ = 800 kg
mass of boat, m₂ = 1400 kg
initial velocity of the fish, u₁ = 12 m/s
initial velocity of the boat, u₂ = 0
let the final velocity of the fish-boat after collision = v
Apply the principle of conservation of linear momentum for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
800 x 12 + 1400 x 0 = v(800 + 1400)
9600 = 2200v
v = 9600/2200
v = 4.36 m/s
Therefore, the velocity of the boats after the collision is 4.36 m/s.
Answer:
5.02 m
Explanation:
Applying the formula of maximum height of a projectile,
H = U²sin²Ф/2g...................... Equation 1
Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.
Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°
Constant: g = 9.8 m/s²
Substitute these values into equation 1
H = (14.021)²sin²45/(2×9.8)
H = 196.5884×0.5/19.6
H = 5.02 m.
Hence the ball goes 5.02 m high
