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koban [17]
3 years ago
15

How is velocity ratio of wheel and axle calculated​

Physics
1 answer:
Rudiy273 years ago
7 0

Answer:

VR = \frac{Radius of the wheel}{Radius of the axle}

Explanation:

Velocity ratio (VR) of a machine is a term that compares the distance moved by effort put into the machine to the distance moved by the load.

A wheel and axle is a device for lifting of a load through a height. It is made up of two circular parts called wheel and axle. Its velocity ratio (VR) can be determined by:

VR = \frac{Radius of the wheel}{Radius of the axle}

For a practical wheel and axle, the diameter of the wheel is greater than the diameter of the axle.

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Will the temperature of a plastic or wooden spoon be more than the temperature of a metal spoon if you put both of these in a cu
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Answer:

no

Explanation:

the metal spoon would be higher temperature

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There is a box on the floor. The following forces are being applied to it:
Sergio [31]

The answer is 6.

Add 4 then divide 2. You should have 18 after you do that, subtract 10 then multiply 20. You then have 6.

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As rotational speed increases, thrust____?
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Increases exponentially is your correct answer
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If the results of a study do not support a hypothesis, it means that experiment failed
butalik [34]
No that is incorrect. If the results do not support the hypothesis than the hypothesis could have been incorrect. Or there is a possibility that the experiment was not done properly. A hypothesis is an educated guess for the results of the experiment. 
6 0
4 years ago
Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4
zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is c_p=1.0\ Btu/lbm.F, and the steam properties as, A-4E:

h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R

Using the energy balance for the system:

\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0
4 years ago
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