A uniform solid disk with a mass of 24.3 kg and a radius of 0.314 m is free to rotate about a frictionless axle. Forces of 90 an
1 answer:
Answer:
α = 9.18 rad/s²
Explanation:
given,
mass of the solid disk = 24.3 Kg
radius of the disk = 0.314 m
Force, F₁ = 90 N
F₂ = 125 N
net force acting on the disk
F = 125 - 90
F = 35 N
Torque
τ = F . r
τ = 35 x 0.314
τ = 11 N.m
we know that
τ = I α
moment of inertia of the solid disk
I = 1.198 kg.m²
now,
11 = 1.198 x α
α = 9.18 rad/s²
the angular acceleration of the disk is equal to 9.18 rad/s²
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Answer:
By using the acceleration formula,
1) 26.2 m/s
The mechanical energy of the divers at any point of their vertical motion is sum of the kinetic energy and the gravitational potential energy:
where
m is the mass of the diver
v is the speed
g = 9.8 m/s^2 is the acceleration due to gravity
h is the height above the water
When the diver is on the cliff, v = 0 (he is at rest), so K=0 and the initial mechanical energy is just potential energy:
where h=35 m is the height of the cliff.
When the diver hits the water above, h = 0, so U=0 and the final mechanical energy is just kinetic energy:
since the total mechanical energy is conserved, we have
And solving the equation for v, we find the speed when they reach the surface of the water:
2) It is converted into thermal energy of the water
When the diver enters the water, he suddenly feels another force acting against the motion of the diver: the resistance of the water. The resistance of the water acts upward, slowing down the diver until he stops.
In this process, the speed of the diver (v) decreases, and therefore the kinetic energy of the diver decreases as well, until it becomes zero.
However, this does not mean that the conservation of energy has been violated. In fact, the kinetic energy of the diver has been converted into thermal energy of the molecules of water surrounding the diver.
The resultant static friction force is equal to 20 N to the left.
Why?
I'm assuming that you forgot to write the question of the exercise, so, I will try to complete it:
"A 50-n crate sits on a horizontal floor where the coefficient of static friction between the crate and the floor is 0.50 . A 20-n force is applied to the crate acting to the right. What is the resulting static friction force acting on the crate?"
So, if we are going to calculate the resulting static friction force, it means that there is no movement, we must remember that the friction coefficient will give us the maximum force before the crate starts to move.
We can calculate the static friction force by using the following formula:
Since the crate is not moving (static), the static friction force acting on the crate will be equal to the applied force.
Calculating we have:
Hence, the static friction force is equal to 20 N to the left (since the applied force is acting to the right)
So,
Since the static friction force is equal to the applied force, the crate does not start to move.
Have a nice day!
<h2>Answer:</h2>
<em>Hello, </em>
<h3><u>QUESTION)</u></h3>Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².
<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>
- Δt = (√2xgxh)/9,8
- Δt = (14√10)/9,8
- Δt ≈ 4,5 s