Car A take a time of 2.55hr and car B take a time of 2.14 hr
We know that distance divide by time is speed
here it is given that car A to reach a gas station a distance 189 km from the school traveling at a speed of 74 km/hr
so speed=distance/time
s=d/t
t=d/s
=189/74
=2.55hr
In case of car B it is given that The distance from the is 199.8km, car b is traveling at a speed of 93 km/hr
s=d/t
t=d/s
=199.8/93
=2.14hr
so from the above given data and the formula we solved and found out the time taken by car A is 2.55h and car B is 2.14h
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The work done by the machine is equal to the product between the force applied and the distance over which the force is applieds, so in this case:

And the power of the machine is equal to the ratio between the work done by the machine and the time taken:
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.
The Answer is Option C
I think...
Sorry If i am wrong...