All categories

4 years ago

Find the volume of the rectangular prism 9 cm 10 cm

Engineering

1 answer:

sergeinik [125]4 years ago

3 0

Answer:

V= 90h cm³ where h is the height of the rectangular prism.

Explanation:

The formula for volume of a rectangular prism is ;

V=l*w*h where;

V=volume in cm³

l= length of prism=10cm

w =width of the prism = 9 cm

Assume the height of the prism as h cm then the volume will be;

V= 10* 9*h

V= 90h cm³

when the value of height of the prism is given, substitute that value with h to get the actual volume of the prism.

You might be interested in

Why is the back-work ratio much higher in the brayton cycle than in the rankine cycle?

zloy xaker [14]

The back-work ratio much higher in the Brayton cycle than in the Rankine cycle because a gas cycle is the Brayton cycle, while a steam cycle is the Rankine cycle. Particularly, the creation of water droplets will be a constraint on the steam turbine's efficiency. Since gas has a bigger specific volume than steam, the compressor will have to work harder while using gas.

<h3>What are modern Brayton engines?</h3>

Even originally Brayton exclusively produced piston engines, modern Brayton engines are virtually invariably of the turbine variety. Brayton engines are also gas turbines.

<h3>What is the ranking cycle?</h3>

A gas cycle is the Brayton cycle, while the Ranking cycle is a steam cycle. The production of water droplets will especially decrease the steam turbine's performance. Gas-powered compressors will have to do more work since gas's specific volume is greater than steam's.

To know more about Rankine cycle, visit: brainly.com/question/13040242

#SPJ4

4 0

1 year ago

You guys are amazing :D

Sloan [31]

Ik i am thank you tho xoxo

3 0

3 years ago

Read 2 more answers

A three-phase wye-connected synchronous generator supplies a network through a transmission line. The network can absorb or deli

Amanda [17]

Answer:

the graph and the answer can be found in the explanation section

Explanation:

Given:

Network rated voltage = 24 kV

Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi

Rn = 0.07 * 8 = 0.56 Ω

Xn = 0.5 * 8 = 4 Ω

If the alternator terminal voltage is equal to network rated voltage will have

Vt = 24 kV/√3 = 13.85 kV/phase

The alternative current is

$I_{a} =\frac{40x10^{6} }{\sqrt{3}*24x10^{3} } =926.2A$

$X_{s} =0.85\frac{13.85}{926.2} =12.7ohm$

The impedance Zn is

$\sqrt{0.56^{2}+4^{2} } =4.03ohm$

The voltage drop is

$I_{a} *Z_{n} =926.2*4.03=3732.58V$

$r_{dc} =\frac{voltage}{2*current} =\frac{13.85}{2*926.2} =7.476ohm$

rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω

The effective armature resistance is

$Z_{s} =\sqrt{R_{a}^{2}+X_{s}^{2} } =\sqrt{8.97^{2}+12.7^{2} } =15.55ohm$

The induced voltage for leading power factor is

$E_{F} ^{2} =OB^{2} +(BC-CD)^{2}$

if cosθ = 0.5

$E_{F} =\sqrt{(13850*0.5)^{2}+(\frac{3741}{2}-926.2*12.7)^{2} } =11937.51V$

if cosθ= 0.6

EF = 12790.8 V

if cosθ = 0.7

EF = 13731.05 V

if cosθ = 0.8

EF = 14741.6 V

if cosθ = 0.9

EF = 15809.02 V

if cosθ = 1

EF = 13975.6 V

The voltage regulation is

$\frac{E_{F}-V_{t} }{V_{t} } *100$

For each value:

if cosθ = 0.5

voltage regulation = -13.8%

if cosθ = 0.6

voltage regulation = -7.6%

if cosθ = 0.7

voltage regulation = -0.85%

if cosθ = 0.8

voltage regulation = 6.4%

if cosθ = 0.9

voltage regulation = 14%

if cosθ = 1

voltage regulation = 0.9%

the graph is shown in the attached image

for 10% of regulation the power factor is 0.81

8 0

3 years ago

The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe carrying water is 2.00 lb/ft^2. Determine the

soldier1979 [14.2K]

Answer:

a) -8 lb / ft^3

b) -70.4 lb / ft^3

c) 54.4 lb / ft^3

Explanation:

Given:

- Diameter of pipe D = 12 in

- Shear stress t = 2.0 lb/ft^2

- y = 62.4 lb / ft^3

Find pressure gradient dP / dx when:

a) x is in horizontal flow direction

b) Vertical flow up

c) vertical flow down

Solution:

- dP / dx as function of shear stress and radial distance r:

(dP - y*L*sin(Q))/ L = 2*t / r

dP / L - y*sin(Q) = 2*t / r

Where dP / L = - dP/dx,

dP / dx = -2*t / r - y*sin(Q)

Where r = D /2 ,

dP / dx = -4*t / D - y*sin(Q)

a) Horizontal Pipe Q = 0

Hence, dP / dx = -4*2 / 1 - 62.4*sin(0)

dP / dx = -8 + 0

dP/dx = -8 lb / ft^3

b) Vertical pipe flow up Q = pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(pi/2)

dP / dx = 8 - 62.4

dP/dx = -70.4 lb / ft^3

c) Vertical flow down Q = -pi/2

Hence, dP / dx = -4*2 / 1 - 62.4*sin(-pi/2)

dP / dx = -8 + 62.4

dP/dx = 54.4 lb / ft^3

7 0

3 years ago

For which situation will structural firefighting protective clothing provide you with adequate protection

jeka94

Missing part

The options are:

A. You will be exposed to splashes of the material.

B. You will have to handle the material.

C. There are high atmospheric concentrations of the material.

D. None of the above.

Answer:

None of the above

Explanation:

Structural firefighting protective clothing can't give adequate protection when one is exposed to splashes of the material. Moreover, it can't help in a situation where one is required to handle the material or when there are high atmospheric concentrations of the material.

SFC may be used when the following conditions are met:

• Contact with splashes of extremely hazardous materials is unlikely.

• Total atmospheric concentrations do not contain high levels of chemicals toxic to the skin.

• There are no adverse effects from chemical exposure to small areas of unprotected skin.

• Live victims who are in need of rescue, such as those found in a terrorist attack.

3 0

3 years ago