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kondaur [170]
3 years ago
11

A 500-m^3 rigid tank is filled with saturated liquid-vapor mixture of water at 200 kPa. If 20% of the mass is liquid and the 80%

of the mass is vapor, the total mass in the tank is: (a) 635 kg (b) 2809 kg (c) 258 kg (d) 500 kg (e) 705 kg
Engineering
1 answer:
guapka [62]3 years ago
3 0

Answer:

(a) 705 kg

Explanation:

it is given that volume =500m^3

pressure = 200 kPa

it is given that 80% mass is vapor so dryness fraction =x=0.8

from the standard table at 200 kPa

v_f=0.001061 \frac{m^3}{kg}

v_g=0.88578\frac{m^3}{kg}

specific volume v=v_f+x\left ( v_g-v_f \right )

v=0.00106+0.8\left ( 0.88578-0.00106 \right )=0.7088\frac{m^3}{kg}

we know that specific\ volume =\frac{volume}{mass}

0.7088=\frac{500}{mass}

mass=\frac{500}{0.7088}=705.456 \ kg

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5 0
2 years ago
Tahir travel twice as far as ahmed, but onley one third as fast. Ahmed starts travel on tuesday at noon at point x to point z 30
shepuryov [24]

Answer:

6:00 pm the next day

Explanation:

Given that

Tahir traveled twice as far as Ahmed. We say,

Ahmed traveled a distance, D

Tahir would travel a distan, 2D

Tahir traveled 1/3 as fast as Ahmed, so we say

Ahmed traveled at a speed, S

Tahir would travel at a speed, S/3

If Ahmed starts travel on tuesday at noon at point x to point z 300km, by 9:00pm,

Time taken by Ahmed to travel is

9:00 pm - 12:00 pm = 9 hours

Ahmed, traveled 300 km in 9 hours, meaning he traveled at 33.3 km in an hour.

Speed, S that Ahmed traveled with is 33.3 km/h

Remember, we stated that Tahir travels at a speed of S/3, that is, The speed of Tahir is

33.3/3 = 11.1 km/h.

300 km would then be traveled in 300 km/11.1 km/h = 27 hours.

Tahir started traveling, 3 hours after Ahmed, that is 12:00 pm + 3:00 hrs = 3:00 pm, and if he's to spend 27 hours on the journey he would reach destination z at 6:00 pm the next day

7 0
3 years ago
O que é necessário se conhecer para a determinação de uma força concentrada em um ponto?
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Answer: its c

Explanation:

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3 years ago
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3 years ago
Read 2 more answers
1. An automobile travels along a straight road at 15.65 m/s through a 11.18 m/s
djyliett [7]

Answer:

a.) Time = 17.13 seconds

b.) 31.88 m

c.) V = 11.18 m/s

d.) V = 7.1 m/s

Explanation:

The initial velocity U of the automobile is 15.65 m/s.

 At the instant that the two vehicles are abreast of each other, the police car starts to pursue the automobile with initial velocity U = 0 at a constant acceleration of 1.96 m/s². Because the police is starting from rest.

For the automobile, let us use first equation of motion

V = U - at.

Acceleration a is negative since it is decelerating with a = 3.05 m/s² . And

V = 0.

Substitute U and a into the formula

0 = 15.65 - 3.05t

15.65 = 3.05t

t = 15.65/3.05

t = 5.13 seconds

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit and applied his brakes and decelerates at 3.05 m/s².

The total time required for the police car to overtake the automobile will be

12 + 5.13 = 17.13 seconds.

b.) Using the third equation of motion formula for the police car at V = 11.18 m/s and a = 1.96 m/s²

V^2 = U^2 + 2aS

Where S = distance travelled.

Substitute V and a into the formula

11.18^2 = 0 + 2 × 1.96 ×S

124.99 = 3.92S

S = 124.99/3.92

S = 31.88 m

c.) The speed of the police car at the time it overtakes the automobile will be in line with the speed zone which is 11.18 m/s

d.) That will be the final velocity V of the automobile car.

We will use third equation of motion to solve that.

V^2 = U^2 + 2as

V^2 = 15.65^2 - 2 × 3.05 × 31.88

V^2 = 244.9225 - 194.468

V = sqrt( 50.4545)

V = 7.1 m/s

8 0
3 years ago
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