Answer:
The statement is true.
Both gravity and centrifugal force act on the Moon which causes it get pulled towards Earth (gravity) and get "flung away" so it doesn't hit us (centrifugal force).
Answer:
0.893 rad/s in the clockwise direction
Explanation:
From the law of conservation of angular momentum,
angular momentum before impact = angular momentum after impact
L₁ = L₂
L₁ = angular momentum of bullet = + 9 kgm²/s (it is positive since the bullet tends to rotate in a clockwise direction from left to right)
L₂ = angular momentum of cylinder and angular momentum of bullet after collision.
L₂ = (I₁ + I₂)ω where I₁ = rotational inertia of cylinder = 1/2MR² where M = mass of cylinder = 5 kg and R = radius of cylinder = 2 m, I₂ = rotational inertia of bullet about axis of cylinder after collision = mR² where m = mass of bullet = 0.02 kg and R = radius of cylinder = 2m and ω = angular velocity of system after collision
So,
L₁ = L₂
L₁ = (I₁ + I₂)ω
ω = L₁/(I₁ + I₂)
ω = L₁/(1/2MR² + mR²)
ω = L₁/(1/2M + m)R²
substituting the values of the variables into the equation, we have
ω = L₁/(1/2M + m)R²
ω = + 9 kgm²/s/(1/2 × 5 kg + 0.02 kg)(2 m)²
ω = + 9 kgm²/s/(2.5 kg + 0.02 kg)(4 m²)
ω = + 9 kgm²/s/(2.52 kg)(4 m²)
ω = +9 kgm²/s/10.08 kgm²
ω = + 0.893 rad/s
The angular velocity of the cylinder bullet system is 0.893 rad/s in the clockwise direction-since it is positive.
Answer:
Explanation:
From the given information:
The initial PE 
= m×g×h
= 5 kg × 9.81 m/s² × 10 m
= 490.5 J
The change in Potential energy P.E of the box is:
ΔP.E = 
ΔP.E = 0 -
ΔP.E = 
If we take a look at conservation of total energy for determining the change in the internal energy of the box;
this can be re-written as:
Here, K.E = 0
Also, 70% goes into raising the internal energy for the box;
Thus,
ΔU = 343.35 J
Thus, the magnitude of the increase is = 343.35 J
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
C, at this moment it is directly 90 degrees to the right and is only travelling downwards
