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3 years ago

Melting, freezing, and boiling are _____________ changes. Question 3 options: A. compound

2 answers:

maks197457 [2]3 years ago

4 0

Melting, freezing, and boiling are PHYSICAL changes. Reason why, you can SEE their PHYSICAL appearance change. Melting, you can see a solid (ice) melt, and turn into a liquid (that's a physical change). Freezing, is when a liquid (possibly water) becomes a solid. It's features change, thus it's a physical change. Boiling, when something is boiled you can see it physically BUBBLE. It heats up, and you can see steam escape from it. This is a physical change, but when you turn it off, the water goes back to normal. But not exactly. When you boil water, you're cleaning and purifying the water. So it become purified drinking water. Also just a tip, when boiling water, make sure the temperature is above 185° F (85° C) so it'll be able to kill all bacteria.
(I wrote this myself, it's my work.)
Good luck with your studies, hope this helps!! ^-^

vfiekz [6]3 years ago

3 0

Melting freezing and boiling are molecular changes

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Answer:

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Explanation:

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3 years ago

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3 years ago

Light waves are traveling through glass. If the wavelength of some part of the light wave is 4 x 10^(-7) m and the frequency of

Ivenika [448]

Wave speed = fλ
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3 0

3 years ago

A 15 m long garden hose han an inner diameter of 2.5 cm. One end is connected to a spigot; 20 C water flows from the other end a

pogonyaev

We can solve this problem using Hagen–Poiseuille equation. Derivation of this equation is a bit complicated so I will just write down the equation.
 $\Delta P= \frac{8\mu QL}{\pi R^4}$
This equation gives you the pressure drop in an incompressible and Newtonian fluid in laminar flow flowing through a long cylindrical pipe of the constant cross section.
L is the length of the cylinder, Q is the volumetric flow rate, R is the radius of the pipe, and $\mu$ is dynamic viscosity.
Dynamic viscosity of water at 20 Celsius is 0.001 PaS.
Now we can calculate the pressure drop:
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7 0

3 years ago

A baseball is hit almost straight up into the air with a speed of 26 m/s . Estimate how high it goes.

Natalka [10]

Answer:

The maximum height of the ball is 34.5 m.

The ball is 5.31 s in the air.

Explanation:

Hi there!

The equations for the height and velocity of the baseball that is hit straight up are as follows:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height of the baseball at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time t.

If we place the origin of the frame of reference at the place where the baseball is hit, then, y0 = 0.

To calculate how high it goes, we have to obtain the time at which the ball is at maximum height. At that point, the velocity is 0. Then using the equation of velocity:

v = v0 + g · t

0 = 26 m/s - 9.8 m/s² · t

-26 m/s / -9.8 m/s² = t

t = 2.65 s

The height at that time will be the maximum height:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

y = 26 m/s · 2.65 s - 1/2 · 9.8 m/s² · (2.65 s)²

y = 34.5 m

The maximum height of the ball is 34.5 m

If it takes the ball 2.65 s to reach the maximum height it will take another 2.65 s to return to the initial position. Then, the time it will be in the air is (2.65 s + 2.65 s) 5.30 s. However, let´s calculate the time it takes the ball to reach the initial position using the equation for height.

At the initial position y = 0. Then:

y = y0 + v0 · t + 1/2 · g · t² (y0 = 0)

0 = 26 m/s · t - 1/2 · 9.8 m/s² · t²

0 = t (26 m/s - 1/2 · 9.8 m/s² · t) (t = 0 when the ball is hit)

0 = 26 m/s - 1/2 · 9.8 m/s² · t

-26 / -4.9 m/s² = t

t = 5.31 s ( the difference with the 5.30 s obtained above is due to rounding the time to 2.65 s).

The ball is 5.31 s in the air.

Have a nice day!

5 0

4 years ago