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Tems11 [23]
2 years ago
6

To lift a load of 100 kg a distance of 1 m an effort of 25 kg must be applied over an inclined plane of length 4 m. What must be

done to lift the load 2 m, using the same effort?
A) Use an inclined plane of length 8 m.
B) Use an inclined plane of length 2 m.
C) Use an inclined plane of length 10 m.
D) Use an inclined plane of length 16 m.
Physics
2 answers:
Troyanec [42]2 years ago
6 0
Using conservation of energy law:-
∑ work in = ∑ work out
and work= force* displacement 
so when we wanted to move a 100kg a distance of 1m 
we multiplied 100*1 = work out
so work in should be equal to 100*g Joules, where g is the acceleration due to gravity.
so workout = 100*g = 25*g *x (divide both sides by 25*g)
x=4m 

by the same way:-
------------------------
work in = 100kg * 2m * g (m/s^2)= work out
so work out = 25*x*g = 200* g (divide both sides by 25*g)
x=8m
insens350 [35]2 years ago
4 0
A) Use an inclined plane of length 8 m.
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The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of
natita [175]

Answer:

3.258 m/s

Explanation:

k = Spring constant = 263 N/m (Assumed, as it is not given)

x = Displacement of spring = 0.7 m (Assumed, as it is not given)

\mu = Coefficient of friction = 0.4

Energy stored in spring is given by

U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J

As the energy in the system is conserved we have

\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s

The speed of the 8 kg block just before collision is 3.258 m/s

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What is the meaning of viscosity?
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Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha
tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

         R_cm = 1 / M   ∑  x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

       R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

      M = m_earth + m_moon

     

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

      M = 6.0535 10₂⁴24 kg

we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

      R_cm = 4.66 10⁶ m

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3 years ago
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