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malfutka [58]
3 years ago
9

An expansion valve does work on 100 g of water. The system is isolated, and all of the work is used to convert the 90°C water in

to water vapor at 110 degrees * C . How much work does the expansion valve do on the water?
Physics
1 answer:
mel-nik [20]3 years ago
7 0

Answer:

The expansion valve removes pressure from the liquid refrigerant to allow expansion or change of state from a liquid to a vapor in the evaporator. The high-pressure liquid refrigerant entering the expansion valve is quite warm. This may be verified by feeling the liquid line at its connection to the expansion valve.

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D. If a dog has a mass of 12 kg, what is its weight on Neptune?<br> 11.7N/kg
steposvetlana [31]

Answer:

133.8 N

Explanation:

Recall that the acceleration of gravity in Neptune is estimated as 11.15 m/s^2

Therefore, the weight of the dog on this planet would be:

Weight = mass x acceleration of gravity = 12 kg x 11.15 m/s^2 = 133.8 N

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3 years ago
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algol [13]

Answer:

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Two forces are applied to a car in an effort to accel-
mestny [16]

Answer:

R=2F

Explanation:

As the forces are in same direction so the resultant force will be:

R=F+F

R=2F

4 0
3 years ago
A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
Mandarinka [93]

Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

6 0
3 years ago
A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the
Nitella [24]

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

4 0
3 years ago
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