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vaieri [72.5K]
3 years ago
14

Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s

Engineering
1 answer:
zlopas [31]3 years ago
5 0

Answer:

4m/s

Explanation:

We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted

Mathematically

Power_{motor} } =\frac{Energy }{time}\

We also know that Power = force x velocity      ..................(i)

The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load

=> power = mg x Velocity........(ii)

While hoisting the load at at constant speed only the potential energy of the mass increases

Thus Potential energy = Mass x g x H...................(iii)

where

g = accleration due to gravity (9.81m/s2)

H = Height to which the load is hoisted  

Equating equations (ii) and (iii) we get

m x g x v = \frac{mgh}{t}

thus we get v = H/t

Applying values we get

v = 6/1.5 = 4m/s

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Answer:

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3 years ago
A plane wall of length L, constant thermal conductivity k and no thermal energy generation undergoes one-dimensional, steady-sta
KIM [24]

Answer:

Temperature distribution is T(x)=\dfrac{q}{k}(L-x)+T

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Explanation:

We know that for no heat generation and at steady state

\dfrac{\partial^2 T}{\partial x^2}=0

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a and are the constant.

Given that heat flux=q

We know that heat flux given as

q=-K\dfrac{dT}{dx}

From above we can say that

a= -\dfrac{q}{K}

Alos given that when x= L temperature is T(L)=T

T= -\dfrac{q}{K}L+b

b=T+\dfrac{q}{K}L

So temperature T(x)

T(x)=-\dfrac{q}{K}x+T+\dfrac{q}{K}L

T(x)=\dfrac{q}{k}(L-x)+T

So temperature distribution is T(x)=\dfrac{q}{k}(L-x)+T

Heat flux=q

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6 0
3 years ago
A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be
8090 [49]

Answer:

The maximum power dissipation of the zener diode 112mV.

Explanation:

The minimum zener current should be:

5 * Iza= 5 * 1=  5 mA.

Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.

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Maximum power dissipated in the diode occours when, IL=0 is

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3 years ago
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Answer: 0.053

Explanation:

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