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kramer
3 years ago
10

A 6157 N piano is to be pushed up a(n) 2.41 m frictionless plank that makes an angle of 21.9 ◦ with the horizontal. Calculate th

e work done in sliding the piano up the plank at a slow constant rate.
Physics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Answer:

Work done, W = 5534.53 J

Explanation:

It is given that,

Force acting on the piano, F = 6157 N

It is pushed up a distance of 2.41 m friction less plank.

Let W is the work done in sliding the piano up the plank at a slow constant rate. It is given by :

W=Fd\ sin\theta

Since, F=F\ sin\theta (in vertical direction)

W=6157\times 2.41\ sin(21.9)

W = 5534.53 J

So, the work done in sliding the piano up the plank is 5534.53 J. Hence, this is the required solution.

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IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
3 years ago
Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet
yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

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The mass flow rate is

m_{2}=0.15\times2

We need to calculate the mass flow rate at reach exit

Using formula of mass

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m_{3}=2-0.15\times2

m_{3}=1.7\ kg/s

We need to calculate the inlet velocity

Using formula of velocity

v=\dfrac{m}{\rho A}

Put the value into the formula

v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}

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Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

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Answer:

The answer is 904,000.

Kinetic energy=1/2mv^2.

1/2×1130×40^2.

1/2×1808000=904,000Joules.

8 0
3 years ago
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Answer: SATURATED

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I believe that the answer to this would be B




Hope this helped
3 0
3 years ago
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