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just olya [345]
3 years ago
6

At each of the designated points, rotate the given vector to indicate the direction of the force exerted by the water on either

the inside of the fishbowl (for points A and B) or the outside of the goldfish (for points C, D, and E).

Physics
1 answer:
GrogVix [38]3 years ago
4 0

Answer:

The direction of the force at A and B is perpendicular to the walls of the container.

The direction of the force at C is down.

The direction of the force in D is up

The direction of the force at E is to the left.

The attached figure shows the forces exerted by the water at points A, B, C, D and E.

Explanation:

The water is in contact with the bowl and with the fish. It exercises at points A, B, C, D and E, but the direction is different from the force.

The fish has a buoyant force on the water and that direction is up. The direction of at point D is up.

The column of water on the fish has a downward force, therefore the direction of the force at point C is down. The water column to the right of the fish has a force to the left, and the direction at point E is to the left.

The water will exert a force on the walls of the container and this force at points A and B is a on the walls of the container.

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What force must the deltoid muscle provide to keep the arm in this position?
ruslelena [56]

Answer:

Deltoid Force, F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.  

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, T_{Deltoid}

2. The torque of the arm, T_{arm}  

Assuming the arm is just being stretched and there is no rotation going on,

                        T_{Deltoid} = 0

                       T_{arm} = 0

       ⇒           T_{Deltoid} = T_{arm}

                  r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}

Where,

r_{d} is radius of the deltoid

F_{d} is the force of the deltiod

\alpha_{d} is the angle of the deltiod

r_{a} is the radius of the arm

F_{a} is the force of the arm , F_{a} = mg  which is the mass of the arm and acceleration due to gravity

\alpha_{a} is the angle of the arm

The force of the deltoid muscle is,

                                 F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}

but F_{a} = mg ,

                ∴            F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}

7 0
3 years ago
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Explanation:

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3 years ago
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a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav
AURORKA [14]
This next statement is a big deal.  It should be up on a board, surrounded
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Hello!

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Have a nice day!


3 0
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