Answer:
Explanation:
Length if the bar is 1m=100cm
The tip of the bar serves as fulcrum
A force of 20N (upward) is applied at the tip of the other end. Then, the force is 100cm from the fulcrum
The crate lid is 2cm from the fulcrum, let the force (downward) acting on the crate be F.
Using moment
Sum of the moments of all forces about any point in the plane must be zero.
Let take moment about the fulcrum
100×20-F×2=0
2000-2F=0
2F=2000
Then, F=1000N
The force acting in the crate lid is 1000N
Option D is correct
I don't understand what you mean by "depth" of the steps. The flat part of the step has a front-to-back dimension, and the 'riser' has a height. I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy. And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground. So something is definitely fishy about the steps.
Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.
In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters. The work [s]he does is (force) x (distance) = <em>662,400 joules. </em>
Answer:
Incomplete question
Check attachment for the given diagram
Explanation:
Given that,
Initial Velocity of drum
u=3m/s
Distance travelled before coming to rest is 6m
Since it comes to rest, then, the final velocity is 0m/s
v=3m/s
Using equation of motion to calculate the linear acceleration or tangential acceleration
v²=u²+2as
0²=3²+2×a×6
0=9+12a
12a=-9
Then, a=-9/12
a=-0.75m/s²
The negative sign shows that the cylinder is decelerating.
Then, a=0.75m/s²
So, using the relationship between linear acceleration and angular acceleration.
a=αr
Where
a is linear acceleration
α is angular acceleration
And r is radius
α=a/r
From the diagram r=250mm=0.25m
Then,
α=0.75/0.25
α =3rad/sec²
The angular acceleration is =3rad/s²
b. Time take to come to rest
Using equation of motion
v=u+at
0=3-0.75t
0.75t=3
Then, t=3/0.75
t=4 secs
The time take to come to rest is 4s
<h3><u>Answer;</u></h3>
= 1.256 m
<h3><u>Explanation;</u></h3>
We can start by finding the spring constant
F = k*y
Therefore; k = F/y = m*g/y
= 0.40kg*9.8m/s^2/(0.76 - 0.41)
= 11.2 N/m
Energy is conserved
Let A be the maximum displacement
Therefore; 1/2*k*A^2 = 1/2*k*(1.20 - 0.41)^2 + 1/2*m*v^2
Thus; A = sqrt((1.20 - 0.55)^2 + m/k*v^2)
= sqrt((1.20 -0.55)^2 + 0.40/9.8*1.6^2)
= 0.846 m
Thus; the length will be 0.41 + 0.846 = 1.256 m
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω
Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂
The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ = (r₁/ r₂)² · ω<span>₁
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ = (4)² · ω<span>₁
= 16 </span>· ω<span>₁
Hence, the protostar will be rotating 16 </span><span>times faster.</span>