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alina1380 [7]
3 years ago
13

State two reasons why the rod is not in equilibrium​

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0

Answer:

For equilibrium the net torque on the rod must be zero. So the 100N forces should on either side of the pivot point i.e. one on the left and one the right, and in same direction.

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A skier moving at 4.75 m/s encounters a long, rough, horizontal patch of snow having a coefficient of kinetic friction of 0.220
disa [49]
First we need to find the acceleration of the skier on the rough patch of snow.
We are only concerned with the horizontal direction, since the skier is moving in this direction, so we can neglect forces that do not act in this direction. So we have only one horizontal force acting on the skier: the frictional force, \mu m g. For Newton's second law, the resultant of the forces acting on the skier must be equal to ma (mass per acceleration), so we can write:
ma=-\mu m g
Where the negative sign is due to the fact the friction is directed against the motion of the skier.
Simplifying and solving, we find the value of the acceleration:
a=-(0.220)(9.81 m/s^2)=-2.16 m/s^2

Now we can use the following relationship to find the distance covered by the skier before stopping, S:
2aS=v_f^2-v_i^2
where v_f=0 is the final speed of the skier and v_i=4.75 m/s is the initial speed. Substituting numbers, we find:
S=- \frac{v_i^2}{2a}=- \frac{(4.75 m/s)^2}{2(-2.16 m/s^2)}=5.23 m
5 0
3 years ago
Two forces (4N and 3N) pull your the left, while a 12N force pulls to the right. What is the net force
Lapatulllka [165]

Answer:

The answer is 11N to the right

Explanation:

Because 4N-3N= 1N

Therefore, 12N-1N=11N

The netforce is 11N to the right, because the greatest force is 12N to the right so it is more likely that the object is being pulled to the right.

8 0
2 years ago
What happens when the falling object reaches its terminal velocity? It will stop _________
lora16 [44]

Answer:

LOL i don't even know

Explanation:

7 0
2 years ago
A 70 ft rope hangs from a helicopter above this room. The rope has a mass per unit length of 2 lb/ft. In order to be rescued fro
Mrac [35]

Answer:

The work done to get you safely away from the test is  2.47 X 10⁴ J.

Explanation:

Given;

length of the rope, L = 70 ft

mass per unit length of the rope, μ = 2 lb/ft

your mass, W = 120 lbs

mass of the 70 ft rope  = 2 lb/ft x 70 ft

                                         = 140 lbs.

Total mass to be pulled to the helicopter, M = 120 lbs  + 140 lbs  

                                                                       = 260 lbs

The work done is calculated from work-energy theorem as follows;

W = Mgh

where;

g is acceleration due gravity = 32.17 ft/s²

h is height the total mass is raised = length of the rope = 70 ft

W = 260 Lb x 32.17 ft/s²  x 70 ft

W = 585494 lb.ft²/s²

1 lb.ft²/s² = 0.0421 J

W = 585494 lb.ft²/s²  = 2.47 X 10⁴ J.

Therefore, the work done to get you safely away from the test is  2.47 X 10⁴ J.

4 0
2 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
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