Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
Answer:
W= 638.1 J
Explanation:
As we know that
Work done is the area of the force and displacement diagram.
W=∫F.dx
W=Work
F=force
dx=Elemental displacement
From the diagram ,area A
A= 60 x 7.09 + 1/2 x 60 x 7.09
A= 638.1 J
So the work W
W= 638.1 J
Answer: 6.21 m/s opposite to Janet
Explanation:
Given
mass of Janet
mass of Sarah
The initial velocity of Janet
The final velocity of Janet after the collision is
the final velocity of Sarah after the collision is
Conserving linear momentum
Negative indicates the direction of velocity is opposite than of Janet
The speed of the second mass after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
<h3>
What are we to consider in equilibrium ?</h3>
Whenever the friction in the pulley is negligible, the two blocks will accelerate at the same magnitude. Also, the tension at both sides will be the same.
Given that a large mass m1=5.75 kg and is attached to a smaller mass m2=3.53 kg by a string and the mass of the pulley and string are negligible compared to the other two masses. Mass 1 is started with an initial downward speed of 2.13 m/s.
The acceleration at which they will both move will be;
a = ( - ) / ( + )
a = (5.75 - 3.53) / (5.75 + 3.53)
a = 2.22 / 9.28
a = 0.24 m/s²
Let us assume that the second mass starts from rest, and the distance covered is the h = 2.47 m
We can use third equation of motion to calculate the speed of mass 2 after it has moved ℎ=2.47 meters.
v² = u² + 2as
since u =0
v² = 2 × 0.24 × 2.47
v² = 1.1856
v = √1.19
v = 1.0888 m/s
Therefore, the speed of mass 2 after it has moved ℎ=2.47 meters will be 1.09 m/s approximately
Learn more about Equilibrium here: brainly.com/question/517289
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Answer: For every force there is a reaction force that is equal in size, but opposite in direction.
Hope it helps:)