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3 years ago

Which person is weightless?

Physics

1 answer:

ahrayia [7]3 years ago

7 0

Answer: Option (b) is the correct answer.

Explanation:

The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.

For example, force of gravity on moon is zero which means any object or person on moon will be weightless.

On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.

Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.

Thus, we can conclude that an astronaut on the Moon is the person who is weightless.

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Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500°C, and 80 m/s and the exit c

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Answer:

a) $\Delta \dot K = 24.570\,kW$ , b) $\dot W_{out} = 12729.15\,kW$ , c) $A_{in} = 0.0136\,m^{2}$

Explanation:

A turbine is a device which works usually in steady state and assumption of being adiabatic means no heat interactions between steam through turbine and surroudings and produce mechanical work from fluid energy. Changes in gravitational energy can be neglected. This system can be modelled after the First Law of Thermodynamics:

$-\dot W_{out} + \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2}) = 0$

a) Change in kinetic energy

$\Delta \dot K = \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2} - v_{out}^{2})$

$\Delta \dot K = \frac{1}{2} \cdot \left(12.6\,\frac{kg}{s} \right) \cdot \left[\left(80\,\frac{m}{s} \right)^{2}-\left(50\,\frac{m}{s} \right)^{2}\right]$

$\Delta \dot K = 24570\,W$

$\Delta \dot K = 24.570\,kW$

b) Power output

$\dot W_{out} = \dot m \cdot (h_{in} - h_{out}) + \frac{1}{2}\cdot \dot m \cdot (v_{in}^{2}-v_{out}^{2})$

$\dot W_{out} = \left(12.6\,\frac{kg}{s}\right)\cdot \left(3446\,\frac{kJ}{kg} - 2437.7\,\frac{kJ}{kg} \right) + 24.570\,kW$

$\dot W_{out} = 12729.15\,kW$

c) Turbine inlet area

Turbine inlet area can be found by using the following expressions:

$\dot V_{in} = \dot m \cdot \nu_{in}$

$\dot V_{in} = \left(12.6\,\frac{kg}{s}\right) \cdot \left(0.086442\,\frac{m^{3}}{kg} \right)$

$\dot V_{in} = 1.089\,\frac{m^{3}}{s}$

$A_{in} = \frac{\dot V_{in}}{v_{in}}$

$A_{in} = \frac{1.089\,\frac{m^{3}}{s} }{80\,\frac{m}{s} }$

$A_{in} = 0.0136\,m^{2}$

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Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.

-- The buoyant force is precisely the missing 30N .

-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

 Weight = (mass) x (gravity)
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The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
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The weight of the displaced water is 30N, and weight = (mass) (gravity).

 30N = (mass of the displaced water) x (9.81 m/s²)

 Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

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Finally, density of the frewium sample = (mass)/(volume)

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================================================

I'm thinking that this must be the hard way to do it,
because I noticed that

 (weight in air) / (buoyant force) = 185N / 30N = 6.1666...

So apparently . . .

 (density of a sample) / (density of water) =

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I never knew that, but it's a good factoid to keep in my tool-box.

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