All categories

3 years ago

EXPERTS/ACE and people that wanna help 4 sure only!

Physics

1 answer:

mario62 [17]3 years ago

3 0

That first one you have selected (3,-3) works in both equations so it's correct.
good job.

you can do this guess and test method with multiple choice answers. If it works in both equations it is the solution. Otherwise use substitution or elimination to combine the two into one equation in only one variable. Then you can solve for the one variable first and use it to solve for the other.

You might be interested in

Hi! Can you help me? :) A satellite's speed is 10,000 m/s. After 1 min, it is 5,000 m/s. What is the satellite's acceleration? T

Kruka [31]

a = ΔV/Δt = (5000-10000)/60 = -500/6 = -83.(3) m/sec^2

7 0

3 years ago

Read 2 more answers

A device has power -2 D . The device is

slavikrds [6]

Answer:

Explanation:

-vesign shows the lens is CONCAVE

f=1/power

7 0

3 years ago

The four stages of listening include hearing, understanding, ________, and responding.

galben [10]

The four stages of listening include hearing, understanding, "Evaluating" and responding.

Hope this helps!

5 0

3 years ago

Which example nest represents balanced forces ???

n200080 [17]

Answer:

an elevator stopped on the third floor a basketball shot into a hoop a sled sliding down a snowy hill a tow truck pulling a car out of a ditch

5 0

3 years ago

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