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MA_775_DIABLO [31]
3 years ago
5

A man lifts various loads with the same lever. The distance of the applied force from the fulcrum is 2.00 m and the distance fro

m the fulcrum to the load is 0.500 m. A graph of resistance force vs. effort force is shown. What is the mechanical advantage of the lever? What is the ideal mechanical advantage of the lever? What is the efficiency of the lever? Show your work.
Physics
2 answers:
lakkis [162]3 years ago
7 0
In your question where as ask the man lift various loads with the same lever and the distance of the applied force from the fulcrum is 2m and the distance from the fulcorm to load is 0.5m.So the following are the answers to your questions:
a.mechanical advantage of the lever is 1
b..<span>ideal mechanical advantage of the lever is 4</span>
c.4
Mandarinka [93]3 years ago
7 0

its 2.5 on edge just got it right .

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A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9
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Answer:

The convection coefficient is 15456.48\ W/m^{2}K

Solution:

Mass flow rate, \dot{m} = 0.4\ kg

Inner diameter of the tube, d = 0.014 m

Fluid density, \rho_{f} = 990\ kg/m^{3}

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, P_{r} = 8.6

Heat flux, \dot{q} = 71,297\ W/m^{2}

Viscosity, \mu = 0.00079\ Ns/m^{2}

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}

Determine the velocity of the fluid inside the tube by mass flow rate:

\dot{m} = \rho_{f}A_{c}v

0.4 = 990\times 1.54\time 10^{- 4}v

v = 2.624 m/s

Determine the Reynold's Number, R_{e}:

R_{e} = \frac{\rho_{f}dv}{\mu}

R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253

Thus it is clear that R_{e} > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}

N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42

Also,

N_{u} = \frac{dh}{K}

where

h = convection coefficient

Now,

292.42 = \frac{0.014\times h}{0.74}

h = 15456.48\ W/m^{2}K

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