<em>To understand the passage between two blades, it is required </em>
<em>to travel the distance of the circumference equivalent to the </em>
<em>segment of the diameter that exists between them,</em>
Where
Ball diameter
Space time
So the angle swept out by either a blade or a space is:
Through the angular velocity
So,
First of all, we have to sort out how this situation is put together. You have 55 kg that's hanging suspended, but it's tied to another 65 kg on a horizontal ledge. Somewhere in there, the rope has to turn a corner. Obviously, it has to go around the edge of the ledge, and we have top assume that there's absolutely no friction against the rope at that point, because we're not told anything different. So we have to treat the edge like a frictionless pulley. And we're also going to ignore the weight of the rope.
OK. The weight of the 55-kg guy who is hanging suspended is (mass) x (gravity) =
(55 x 9.8) = 539 newtons. That force soaks upward through the rope, makes the turn
at the edge, and is exerted horizontally against the 65-kg guy on the ledge.
Hopefully, the suspended guy can hold on for a few minutes longer, while we analyze
the forces around the heavier guy up on the ledge.
The weight of the guy on the ledge is (mass) x (gravity) = (65 x 9.8) = 637 newtons.
That's his weight, pointing downward, against the ledge.
As his boots slip along the ledge, the friction force against his motion is
(his weight) x (the coefficient of kinetic friction between him and the ledge) =
(637 newtons) x (0.45) = 286.65 newtons.
The man on the ledge has the rope pulling him toward the edge with 539N of force,
and 286.65N of friction force holding him back. You can see that he's slipping toward
the edge, because the friction force isn't enough to hold him.
The net force on him is (539N forward) + (286.65N backward) = 252.35N forward.
Since the man on the ledge has a net force pulling him forward toward the edge,
he accelerates in that direction.
Force = (mass) x (acceleration)
Acceleration = (force) / (mass) = (252.35N) / (65kg) = <em><u>3.88 meters per second²</u></em>
He's sliding toward the edge with an acceleration of about 0.396 G ... his speed is increasing about 39 or 40% as fast as it will after he falls over the edge, and the both of them proceed toward their ultimate and apparently unavoidable 'splut' below.
I've totally terrified myself answering this one.
Answer:
A) I_total = 16 m, B) I_total = 8 m, C) I_total = 8 m, D) I_total = 8 m
Explanation:
The moment of inertia is a scalar quantity, therefore the total moment of inertia
I_total = I₁ + I₂ + I₃ + I₄
the moment of inertia of a point mass with respect to an axis of rotation
I = m r²
Let's apply this to our case
A) Rotation axis at the origin
I₁ = m 0 = 0
for the second masses, we find the distance using the Pythagorean theorem
r =
r = 2 √2
I₂ = m (2 √2) ²
I₂ = 8 m
I₃ = m 2² = 4 m
I₄ = m 2² = 4 m
we substitute
I_total = 0 + 8m + 4m + 4m
I_total = 16 m
B) axis of rotation in the center of the square
let's find the distance to any mass
r =
r = √2
I₁ = m 2
I₂ = m 2
i₃ = m 3
I₄ = m 4
we substitute
I_total = 4 (2m)
I_total = 8 m
C) axis of rotation is the x axis
I₁ = 0
I₂ = m 2² = 4 m
I₃ = m 2² = 4 m
I₄ = 0
I_total = 8 m
D) axis of rotation is the y-axis
I₁ = 0
I₂ = 4m
I₃ = 0
I₄ = 4 m
I_total = 8 m
Answer:
A. 243 N
Explanation:
Friction is the force that opposes the relative motion between systems that are in contact.
This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.
First find the normal force which is the force that would point directly upwards to support weight of the block.
Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.
Given m=40 kg and g=9.80 m/s²
N force=40×9.80 =392N
Then find the force of friction which is given by the formula;
<em>F=μN where μ is friction coefficient for the oak chest and N is the normal force on the chest</em>
Given <em>μ</em>=0.620 and N force = 392 N then it will be;
F=0.620× 392 =243.04 N
Answer : 243 N
Explanation:
As a tornado or other storm system passes over a building, low pressure can tug a roof upward. When those forces surpass the force exerted by the weight of the roof, the structure flies up and is swept away by wind currents..