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Volgvan
4 years ago
12

Determine the heat energy required to vaporize 13.9 grams of liquid water at 100° C. O 2,006 cal O 47.8 cal O 7,506 cal O 24.9 c

al
Physics
1 answer:
Harlamova29_29 [7]4 years ago
6 0

Answer:

7506 cal

Explanation:

Given: Mass of water = m = 13.9 grams.

Heat energy required to determine the conversion of 13.9 grams of water to steam or vapor is m L,

where L = Latent heat of vaporization of water=540 cal/g.

Q = mL = (13.9)(540) = 7,506 cal.

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HELP !! Maura is deciding which hose to use to water her outdoor plants. Maura noticed that the water coming out of her garden h
MA_775_DIABLO [31]
THE GREEN HOSE:
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4          (1)

Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4                    (2)

Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4                     (3)

To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8            (4)

Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0             

Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).

From (3), obtain
a = -4/4.142² = -0.2332

The equation for the green hose is
y = 0.2332(x - 4.142)² + 4

THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.

A graph of y(x) for both hoses is shown in the attached figure.

Answers:
a. The red hose will throw the water higher. 

b. The equation for the green hose is
     y = -0.2332(x - 4.124)² + 4,
     with the origin at a height of 4 feet above ground level.

c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
     The corresponding range is -4 ≤ y ≤ 4 feet.


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