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3 years ago

Water boils to power a steam engine. Which statement best describes the changes in the water as it boils?

Physics

2 answers:

hammer [34]3 years ago

8 0

Option B

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crimeas [40]3 years ago

7 0

Answer:

C. The potential energy of the particles increases as the intermolecular forces are overcome

Explanation:

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago

Question 9 of 10<br> How many carbon (C) atoms are in a molecule of CH,O?

Juli2301 [7.4K]

Answer:

1 molecule not shure

Explanation:

6 0

3 years ago

Read 2 more answers

Which statement about projectile motion is true?

Svetllana [295]

C. The range of a projectile increases with an increase in the angle of launch.

3 0

3 years ago

Read 2 more answers

A red cross helicopter takes off from headquarters and flies 120 km at 70 degrees south of west. There it drops off some relief

fiasKO [112]

Answer:

130 km at 35.38 degrees north of east

Explanation:

Suppose the HQ is at the origin (x = 0, y = 0)

So the coordinates of the helicopter after the 1st flight is

$x_1 = -120cos70^o = -41.04 km$

$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

3 0

3 years ago

a 75-kg refrigerator is located on the 70th floor of a skyscraper (300 meters above ground). what is the potential energy of the

mart [117]

Here, you can calculate it's potential energy with respect to ground.
We know, U = mgh
Here, m = 75 Kg
g = 9.8 m/s² [ constant value for earth system ]
h = 300 m

Substitute their values into the expression:
U = 75 × 9.8 × 300
U = 220500 J

In short, Your Final Answer would be 220,500 J

Hope this helps!

8 0

3 years ago