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3 years ago

You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

Physics

2 answers:

enyata [817]3 years ago

6 0

Less than true weight

Darina [25.2K]3 years ago

5 0

Answer: A

Explanation: Your body gets pulled down, causing ur body to have more weight but not mass.

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Of the following which is the largest body?

ankoles [38]

Answer:

Ganymede is the largest body

Explanation:

it is the satellite of jupiter

3 0

3 years ago

Read 2 more answers

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

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

Download pdf

7 0

3 years ago

A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min

Marianna [84]

Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²

   = (1/2) (9.8 m/s²) (95 sec)²

   = (4.9 m/s²) (9,025 sec²)

   = 44,222.5 meters .

The depth of the mine shaft is five times the height of Mt. Everest !

6 0

3 years ago

As the object's particles move faster, the thermal energy increases, causing the _________ to also increase. *

ryzh [129]

Temperature that will be my answer number 1

7 0

3 years ago

Read 2 more answers

On a caterpillars map all distances are marked in kilometers . The caterpillars map shows the distance between two milkweed plan

frutty [35]

Answer:

The distance in kilometers is 4012 × $10^{-6}$ km.

Explanation:

We know that the conversion of 1 millimeters is equal to $10^{-3}$ meter. And then the conversion of 1 meter is equal to $10^{-3}$ km. Then the conversion of 1 millimeter to km will be

1 mm = $10^{-3}$ m

1 m = $10^{-3}$ km

So, 1 mm = $10^{-3}$ × $10^{-3}$ km = $10^{-6}$ km.

As here the the distance is 4012 mm, then the distance in km will be

4012 mm = 4012 × $10^{-6}$ km.

So the distance is 4012 × $10^{-6}$ km.

5 0

3 years ago