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:
In coin card experiment smooth card is used so that the card can slide easily from glass
Answer:
t = 1.02 s
Explanation:
The computation of the time required is shown below:
The package speed for belt is
= 3 - 1
= 2 m/s
Moreover, the decelerative force would be acted on the block i.e u.m.g
So, the decelerative produced
= 0.2 × 9.81
= 1.962 m/s^2
And, final velocity = 0
v = u - at
here
V = 0 = final velocity
u = 2 m/s
so,
0 = 2 - 1.962 × t
t = 1.02 s
Answer:
P₂ = 138.88 10³ Pa
Explanation:
This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation
Let's start by looking for the top speed
Q = A₁ v₁ = A₂ v₂
We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)
v₂ = A₁ / A₂ v₁
They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)
v₂ = 2 1.20
v₂ = 2.40 m / s
Now let's write the Bernoulli equation
P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
Let's clear the pressure at point 2
P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)
we put our reference system at the lowest point
y₁ - y₂ = -20 cm
Let's calculate
P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)
P₂ = 143 103 - 2,160 103 - 1,960 103
P₂ = 138.88 10³ Pa
The following that are not true are E and D
