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3 years ago

Can someone please help me!

Physics

1 answer:

Dimas [21]3 years ago

4 0

Hello there!

The correct answer is 7.24

See picture below for the steps!

As always, I am here to help!

You might be interested in

How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?

Blizzard [7]

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

6 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

Your friend is bragging about his motorcycle. He claims that it can go from a stopped position to 50 miles per hour in three sec

andrew-mc [135]

Acceleration because it is at a rise in speed. The formula for acceleration is Speed/Time.

5 0

3 years ago

Which 3 components must be used in order for a light bulb to light?

crimeas [40]

Answer:

a,e,d

Explanation:

i got it right when i was doing study island

6 0

3 years ago

A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle

tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

$-h= vsinA\times t-gt^2/2$

putting values h=15 m, v=0.8

$-15 = 0.8vt - 4.9t^2$ ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

$-15 = 0.8\times40/0.6 - 4.9t^2$

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = $-H =v sinA t - gt^2/2$

⇒ $4.9t^2 - 8.9\times0.8t - 100 = 0$

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0

3 years ago