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svlad2 [7]
3 years ago
14

An inclined plane of length 12m is used to pull a load of 3000N to height of 3m by applying an effort of 1000N. Calculate its MA

, VR and efficiency. Also, find the work done in the machine
Physics
1 answer:
vovikov84 [41]3 years ago
4 0

Answer:

MA  = 4

VR = 3

e = 0.75 or 75%

work = 12000 [J]

Explanation:

The mechanical advantage in an inclined plane is defined as the relationship between the length of the inclined plane and the height of the plane with respect to the ground.

MA = L/h

MA = 12/3 = 4

Now the real advantage can be calculated as the relationship between the weight of the body and the force being applied.

VR = W/F

VR = 3000/1000

VR = 3

Efficiency can be calculated as the relationship between the real mechanical advantage over the ideal mechanical advantage.

e = VR/MA

e = 3/4

e = 0.75 or 75%

The work is defined as the product of the force by the distance, therefore we have:

work = F*d

work = 1000*12

work = 12000 [J]

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Answer:

option A

Explanation:

given,                                              

height of the drop of stone = 9.44 m

speed of the stone = ?                          

As the stone is dropped the energy of the stone will be conserved.

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3 0
3 years ago
Dos cargas puntuales q1 = −50μC y q2 = +30μC se encuentran
alina1380 [7]

Answer:

Datos:

q1 = -50 μC = 50*10^{-6}

q2 = +30 μC = 30*10^{-6}

F = 10 N

a) x si la <em>F = 10N</em>

Aplicando la Ley de Coulomb:

x = \sqrt\frac{k0 * q1 *q2}{F} = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{10} = 1,162m

b) x si la <em>F = 20 N</em>

x=<em> </em>\sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{20}<em> </em>= 0,822m

c)x si la <em>F = 50 N</em>

x = \sqrt\frac{(9*10^{9} )*(50*10^{-6})*(30*10^{-6})}{50} = 0,520m

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