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damaskus [11]
3 years ago
7

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency

percentage of the machine
Physics
2 answers:
I am Lyosha [343]3 years ago
8 0

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

Explanation:

A load of 20N is lifted through a height of 8m by a machine.If the effort of 80N moves a distance of 6m.Calculate the efficiency percentage of the machine

bekas [8.4K]3 years ago
6 0

Answer:

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<em>Join / LoginQuestionIn a lifting machine</em><em> </em><em>an effort of 500 N is to be moved by a distance of 20 m to raise a load of 10000 N by a distance of 0.8 m</em><em> </em><em> Determine the velocity ratio and mechanical advantage</em>

<em> Determine the velocity ratio and mechanical advantageA</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20Easy</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in App</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolution</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolution</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by Toppr</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option is</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8m</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEd</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820VR=25</em>

<em> Determine the velocity ratio and mechanical advantageA10 and 35B20 and 35C10 and 25D25 and 20EasyOpen in AppSolutionVerified by TopprCorrect option isD25 and 20Distance moved by effort is 20m,that of load is 0.8mVR=LdEdVR=0.820VR=25the load is 10,000N and effort=500N</em>

<em>MA=effortdloadd</em>

<em>MA=effortdloaddMA=50010000</em>

<em>MA=effortdloaddMA=50010000MA=20</em>

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un conductor de plata (p=1,6x10°-6ohm x m) tiene una seccion de 5x10°-6 m°2 y una longitud de 110 m . calcular la resistencia
Alexus [3.1K]

Responder:

35,2 ohm.

Explicación:

Dado:

La resistencia específica del conductor es,

La longitud del conductor es,

El área de la sección transversal del conductor es,

Sabemos que la resistencia de un conductor es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

Por lo tanto, la resistencia se puede expresar como:

R=\frac{\rho\times l}{A}

Ahora, conecte los valores dados y resuelva para 'R'. Esto da,

R=\frac{1.6\times 10^{-6}\ ohm\cdot m\times 110\ m}{5\times 10^{-6}\ m^2}\\\\R=35.2\ ohm

Por lo tanto, la resistencia del conductor es de 35,2 ohm.

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A machine
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Answer:

Power_input = 85.71 [W]

Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

W = F*d

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

W = (80*10)*3\\W = 2400 [J]

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

P=\frac{W}{t}

where:

P = power [W] (units of watts)

W = work [J]

t = time = 40 [s]

P = 2400/40\\P = 60 [W]

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]

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Answer:

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Explanation:

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A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit
likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

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So mass density \mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m

Tension in the wire T = 25 N

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And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

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From the relation we can see that power P\ \propto\ A^2

So if amplitude is halved then power will be \frac{1}{4} times

So power will be equal to \frac{0.2023}{2}=0.0505watt

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