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frutty [35]
3 years ago
12

As you found out in the previous part, bernoulli's equation tells us that a fluid element that flows through a flow tube with de

creasing cross section moves toward a region of lower pressure. physically, the pressure drop experienced by the fluid element between points 1 and 2 acts on the fluid element as a net force that causes the fluid to __________. hints
Physics
1 answer:
8090 [49]3 years ago
8 0
This causes the fluid to increase its speed. Bernoulli's principle tells us that an increase in the speed of a fluid happens at the same time with a reduction in pressure or a reduction in the fluid's potential energy. This necessitates that the amount of kinetic energy, potential energy and internal energy stays persistent.
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None because Neon is already stable
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The critical angle for a beam of light passing from water into air is 48.8°. this means that all light rays with an angle of inc
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totally internally reflected

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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}}

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<span>V=<span>(√</span><span>V2x</span>+<span>V2y</span>) 
</span>
and angle make from +ve x axis is

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0.006<357<700.003<6010<9256.0<9520.00

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