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Darina [25.2K]
3 years ago
13

Identify and sketch all the external forces acting on the chair. Because the chair can be represented as a point particle of mas

s m, draw the forces with their tails centered on the black dot in the middle of the chair. Be certain to draw your forces so that they have the correct orientation. Draw the vectors starting at the black dot. The location and orientation of the vectors will be graded. The length of the vectors will not be graded.
Physics
1 answer:
Monica [59]3 years ago
5 0

Answer:

y axis normal (N) and the weight (W)

x axis  pplied force (F) and friction force (fr)

Explanation:

If we have a chair on a horizontal surface, the normal (N) and the weight (W) of the body act on the vertical axis.

On the x axis, the applied force (F) acts in the direction of movement and the friction force (fr) in the opposite direction of movement.

In this exercise we assume that the body tends to move to the right, all the forces can be seen in the adjoint

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A box is being dragged with a horizontal force of 65 N for 12 meters. If there is a force of friction acting on it
asambeis [7]

Answer:

A. 780 J

B. 120 J

C. 660 J

Explanation:

From the question given above the following data were obtained:

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

A. Determination of the work done by the dragging force.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Workdone (Wd) by dragging force =?

Wd = Fₔ × s

Wd = 65 × 12

Wd = 780 J

Therefore, the work done by the dragging force is 780 J

B. Determination of the work done by friction.

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Workdone (Wd) by friction =?

Wd = Fբ × s

Wd = 10 × 12

Wd = 120 J

Therefore, the work done by friction is 120 J

C. Determination of the net work done on the box.

Dragging force (Fₔ) = 65 N

Distance (s) = 12 m

Force of friction (Fբ) = 10 N

Net work done (Wd) =?

Next, we shall determine the net force acting on the box. This can be obtained as follow:

Dragging force (Fₔ) = 65 N

Force of friction (Fբ) = 10 N

Net force (Fₙ) =?

Fₙ = Fₔ – Fբ

Fₙ = 65 – 10

Fₙ = 55 N

Thus, the net force acting on the box is 55 N

Finally, we shall determine the net work done on the box as follow:

Distance (s) = 12 m

Net force (Fₙ) = 55 N

Net work done (Wd) =?

Wd = Fₙ × s

Wd = 55 × 12

Wd = 660 J

Therefore, the net work done on the box is 660 J

4 0
3 years ago
A disc with a mass of 1 kg moves horizontally to the right with a speed of 7 m/s on a table with negligible friction when it col
Elodia [21]

Answer:

1.6 m/s

Explanation:

First you need to find the momentums of each disc by multiplying their velocities with mass.

disc 1: 7*1= 7 kg m/s

disc 2: 1*9= 9 kg m/s

Second, you need to find the total momentum of the system by adding the momentums of each sphere.

9+7= 16 kg m/s

Because momentum is conserved, this is equal to the momentum of the composite body.

Finally, to find the composite body's velocity, divide its total momentum by its mass. This is because mass*velocity=momentum

16/10=1.6

The velocity of the composite body is 1.6 m/s.

7 0
2 years ago
The four fundamental types of organic molecules are nucleic acids, proteins, lipids, and ________.
notka56 [123]
The other of the four organic molecules is carbohydrates. 
All four of them are:
Nucleic Acids (DNA and RNA stuff)
Proteins
Lipids
& Carbohydrates
4 0
3 years ago
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Water can form large dewdrops in nature how would droplets made of vegetable oil instead of water be different
Natasha2012 [34]

Answer:d

Explanation: oil would form droplets but only tiny ones because it’s surface tension is lower than that of water

3 0
2 years ago
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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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