that statement is true
a Third class lever applied when the effort place between the load and the fulcrum.
For example, in a forearm serve
Fulcrum : The elbow
Effort : The effort that putted by the biceps muscle
Load : The arm
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2
Hi there!
We know that:
U = Potential Energy (J)
K = Kinetic Energy (J)
E = Total Energy (J)
At 10m, the total amount of energy is equivalent to:
U + K = 50 + 50 = 100 J
To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:
100J = mgh
We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².
50J = 10(10)m
m = 1/2 kg
Now, solve for height given that E = 100 J:
100J = 1/2(10)h
100J = 5h
<u>h = 20 meters</u>
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:
Angular momentum L = Iω, where
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>
<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>
<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>
<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>
<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
<span>≈ 2.72 kg*m²/s</span>
