Answer:
w = 4,786 rad / s
, f = 0.76176 Hz
Explanation:
For this problem let's use the concept of angular momentum
L = I w
The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved
Initial Before sticking
L₀ = 0 + I₂ w₂
Final after coupling
= (I₁ + I₂) w
The moments of inertia of a disk with an axis of rotation in its center are
I = ½ M R²
How the moment is preserved
L₀ =
I₂ w₂ = (I₁ + I₂) w
w = w₂ I₂ / (I₁ + I₂)
Let's reduce the units to the SI System
d₁ = 60 cm = 0.60 m
d₂ = 40 cm = 0.40 m
f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz
Angular velocity and frequency are related.
w₂ = 2 π f₂
w₂ = 2π 3.33
w₂ = 20.94 rad / s
Let's replace
w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)
w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)
Let's calculate
w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)
w = 20.94 1.28 / 5.6
w = 4,786 rad / s
Angular velocity and frequency are related.
w = 2π f
f = w / 2π
f = 4.786 / 2π
f = 0.76176 Hz
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Lifting a mass to a height, you give it gravitational potential energy of
(mass) x (gravity) x (height) joules.
To give it that much energy, that's how much work you do on it.
If 2,000 kg gets lifted to 1.25 meters off the ground, its potential energy is
(2,000) x (9.8) x (1.25) = 24,500 joules.
If you do it in 1 hour (3,600 seconds), then the average power is
(24,500 joules) / (3,600 seconds) = 6.8 watts.
None of these figures depends on whether the load gets lifted all at once,
or one shovel at a time, or one flake at a time.
But this certainly is NOT all the work you do. When you get a shovelful
of snow 1.25 meters off the ground, you don't drop it and walk away, and
it doesn't just float there. You typically toss it, away from where it was laying
and over onto a pile in a place where you don't care if there's a pile of snow
there. In order to toss it, you give it some kinetic energy, so that it'll continue
to sail over to the pile when it leaves the shovel. All of that kinetic energy
must also come from work that you do ... nobody else is going to take it
from you and toss it onto the pile.
Answer:
Explanation:
<u>Motion in The Plane</u>
When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.
The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of 
and 
as the initial speed and angle, then we have
If we want to know the maximum height reached by the object, we find the value of t when 
becomes zero, because the object stops going up and starts going down
Solving for t
Then we replace that value into y, to find the maximum height
Operating and simplifying
We have
The maximum height is
