All categories

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

You might be interested in

Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

3 years ago

A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m

saw5 [17]

Answer:

Ep = 3924 [J]

Explanation:

To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

$E_{p}=m*g*h\\$