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Brilliant_brown [7]
3 years ago
5

The solid steel shaft AC has a diameter of 25 mm and is supported by smooth bearings at D and E. It is coupled to a motor at C ,

which delivers 3 kW of power to the shaft while it is turning at 50rev/s. Gears A and B remove 1 kW and 2 kW, respectively. The shaft is free to turn in its support bearings D and E.

Physics
1 answer:
Degger [83]3 years ago
4 0

Answer:

Maximum shear stress in region AB=1.04 MPa

Maximum shear stress in region BC=3.11 MPa

Explanation:

The explanation is attached in the attachments.

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Water flows straight down from an open faucet. The cross-sectional area of the faucet is 2.4 × 10-4m2 and the speed of the water
Ksenya-84 [330]

To solve this problem it is necessary to apply the continuity equations in the fluid and the kinematic equation for the description of the displacement, velocity and acceleration.

By definition the movement of the Fluid under the terms of Speed, acceleration and displacement is,

v_2^2 = v_1^2 + 2gh

Where,

V_i = Velocity in each state

g= Gravity

h = Height

Our values are given as,

A_1 = 2.4*10^{-4} m^2

v_1 = 0.8 m/s

h = 0.11m

Replacing at the kinetic equation to find V_2 we have,

v_2 = \sqrt{v_1^2 + 2gh}

v_2 = \sqrt{(0.8 m/s)^2 + 2(9.80 m/s2)(0.11 m)}

v_2= 1.67 m/s

Applying the concepts of continuity,

A_1v_1 = A_2v_2

We need to find A_2 then,

A_2= \frac{A_1v_1 }{v_2}

So the cross sectional area of the water stream at a point 0.11 m below the faucet is

A_2= \frac{A_1v_1 }{v_2}

A_2= \frac{(2.4*10^{-4})(0.8)}{(1.67)}

A_2= 1.14*10^{-4} m2

Therefore the cross-sectional area of the water stream at a point 0.11 m below the faucet is 1.14*10^{-4} m2

8 0
3 years ago
Solid barium sulfate is placed into a beaker to form a saturated solution of barium sulfate. the solution has a barium concentra
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Using the count data and observational data you acquired, calculate the number of CFUs in the original sample
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2 years ago
A 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s. (a) calculate the mechanical work done by the sprinter d
mezya [45]

The mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.

Work in physics is the energy that is transferred to or from an item when a force is applied along a displacement. It is frequently described in its most basic form as the result of force and displacement.

The quantity of energy moved or transformed per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.. A scalar quantity is power.

Given 75-kg sprinter accelerates from rest to a speed of 11.0 m/s in 5.0 s.

So let,

m = 75 kg

v = 11.0 m/s

t = 5.0 s

So the mechanical work done by the sprinter during this time will be as follow:

W = 0.5 mv²

W = 0.5 (75)(11)²

W = 4537.5 J

The average power the sprinter must generate will be as follow:

Power(P) = W / t

P =  4537.5/5

P = 907.5 W

Only 25% is absorbed. So, the sprinter only absorbed 226.875 J per second which is equal to 54.20 calories per second.

Hence   mechanical work done by the sprinter during this time will be 4537.5 J , the average power the sprinter must generate will be 907.5 W and if the sprinter converts food energy to mechanical energy with an efficiency of 25% then he will be burning calories at 54.20 calories per second.

Learn more about mechanical power here:

brainly.com/question/25573309

#SPJ10

8 0
1 year ago
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