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svetlana [45]
2 years ago
5

An insulated box containing helium gas falls from a balloon 4.5 km above the earth's surface. calculate the temperature rise in

℃ of the helium when box hits the ground. A. 15.2 B. 12.6 C. 25.3 D. 14.1
Engineering
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

The rise in temperature of helium gas is 14.25 K.

Explanation:

Given that

H= 4.5 km

The box is insulated ,at 4.5 km height helium gas have potential energy and when it come down then this potential energy will convert into the internal energy of helium gas.

So by heat balance

mgh=mC_v\Delta T

gh=C_v\Delta T

C_v=3.15\frac{KJ}{kg.K} for helium gas

Now by putting the values

gh=C_v\Delta T

10 x 4.5 =3.15 x ΔT

ΔT=14.25 K

So the rise in temperature of helium gas is 14.25 K.

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When leveraging Artificial Intelligence (AI) in today's business landscape, which statement represents the data leveraged for so
adoni [48]

Answer:

Some AI could end up taking the jobs of many humans.

Explanation:

Their intelligence exponentially grows and human intelligence grows gradually, causing more ideas and better work efficiency from AI.

8 0
2 years ago
A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in
RideAnS [48]

To resolve this problem we have,

R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm

F_{f2} is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa

After that, we can calculate the flexural strenght for the square cross section using the previously value.

\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN

6 0
2 years ago
Poles are values of Laplace transform variable, s, that make denominator of transfer function zero. Zeros are values of Laplace
Ostrovityanka [42]

Answer:

Zero 1 = -1

Zero 2 = -3

Pole 1 = 0

Pole 2 = -2

Pole 3 = -4

Pole 4 = -6

Gain = 4

Explanation:

For any given transfer function, the general form is given as

T.F = k [N(s)] ÷ [D(s)]

where k = gain of the transfer function

N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.

D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.

k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]

it is evident that

Gain = k = 4

N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)

= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)

The zeros are -1 and -3

D(s) = s⁴ + 12s³ + 44s² + 48s

= s(s³ + 12s² + 44s + 48)

= s(s + 2)(s + 4)(s + 6)

The roots are then, 0, -2, -4 and -6.

Hope this Helps!!!

3 0
2 years ago
Michelle is the general manager of a power plant. This morning, she will meet with city officials to discuss environmental issue
Irina-Kira [14]

Answer:

interpersonal.

Explanation:

Out of all the activities performed by Michelle, three activities involves the interpersonal skills.

1. Meeting with city officials

2. Meeting with section managers

3. Handling the complaint filed by an employee

All these activities involves interpersonal skills. Hence, we can say that she had spent her most of the day by using the interpersonal skills.

6 0
2 years ago
A double-acting duplex pump with 6.5-in. liners, 2.5-in. rods, and 18-in. strokes was operated at 3,000 psig and 20 cycles/min.
Stella [2.4K]

Answer:

Pump factor = Fp =  7.854 gal/cycle

Ev = 82.00 %

P_H = 183.29 hp

Explanation:

Given data:

Dimension of duplex pump

6.5 inch liner  

2.5 inch rod

18 inch strokes

Pressure 3000 psig

Pit dimension

7 ft wide

20 ft long

Ls = 18 inch

Velocity = (18)/10

volumetric efficiency is given as E_v = (Actual flow rate)/(Theortical flow rate) * 100

we know that flow rate is given as = Area * velocity

Theoritical flow rate = \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times N

Ev = \frac{7\times 12 \times 20\times 12\times 12 \times \frac{18}{10} inch^3/min}{\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 20}

Ev = 82.00 %

Pump factor Fp = = \frac{\pi}{2}\times Ls(2d_l^2 - d_r^2)\times Ev

Fp =\frac{\pi}{2} \times 18 (2\times 6.5^2 -2.5^2) \times 0.82

Fp = 1814.22 in^3/cyl

Fp =  7.854 gal/cycle

Flow rate q = NFp = 20 \times 7.854 = 157.08 gal/min

Power Ph = \frac{\DeltaP q}{1714} = \frac{3000 \times 157.08}{1714} = 274.93 hp

6 0
2 years ago
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