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Sliva [168]
3 years ago
7

What is a description of compounds?

Physics
2 answers:
elena-14-01-66 [18.8K]3 years ago
8 0
Compounds are elements that are chemically combined, like water for example (it’s both hydrogen and oxygen.)
seraphim [82]3 years ago
4 0

Answer:

Hey!

Compounds are substances that compose of many identical molecules and composed of atoms from more than one element held together by chemical bonds.

Explanation:

For Example: SALT ( Sodium + Chlorine ⇒ SODIUM CHLORIDE

I HOPE THIS HELPED!

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1.) an object moves along the x axis, subject to the potential energy shown. The object has a mass of 1.1kg and starts at rest a
MrRa [10]
If I'm not wrong #1 should be C
4 0
3 years ago
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jek_recluse [69]

Answer:

HII CAN I HAVE BRAINLIST PLEASE I NEED ONE MORE

Explanation:

5 0
3 years ago
A tennis ball is dropped from a height of 10.0 m. It rebounds off the floor and comes up to a height of only 4.00 m on its first
Dominik [7]

Answer:

a) V=14.01 m/s

b) V=8.86 \, m/s

c)t = 2.33s

Explanation:

Our most valuable tool in solving this problem will be the conservation of mechanical energy:

E_m = E_k +E_p

That is, mechanical energy is equal to the sum of potential and kinetic energy, and  the value of this E_m mechanical energy will remain constant. (as long as there is no dissipation)

For a point particle, we have that kinetic energy is:

E_k = \frac{1}{2} m \, V^2

Where m is the mass, and V is the particle's velocity,

Potential energy on the other hand is:

E_p= m\, g\, h

where g is the acceleration due to gravity (g=9.81 \, m/s^2) and h is the height of the particle. How do we define the height? It's a bit of an arbitrary definition, but we just need to define a point for which h=0, a "floor". conveniently we pick the actual floor as our reference height, but it could be any point whatsoever.

Let's calculate  the mechanical energy just before the ball is dropped:

As we drop the ball, speed must be initially zero, and the height from which we drop it is 10 meters, therefore:

E_m = \frac{1}{2}m\,0^2+ mg\cdot 10 \,m\\E_m=mg\cdot10 \, m

That's it, the actual value of m is not important now, as we will see.

Now, what's the potential energy at the bottom? Let's see:

At the bottom, just before we hit the floor, the ball is no longer static, it has a velocity V that we want to calculate, on the other hand, it's height is zero! therefore we set h=0

E_m = \frac{1}{2}m\,V^2+ mg\cdot 0\\\\E_m = \frac{1}{2}m\,V^2

So, at the bottom, all the energy is kinetic, while at the top all the energy is potential, but these energies are the same! Because of conservation of mechanical energy. Thus we can set one equal to the other:

E_m = \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\\\ \frac{1}{2}m\,V^2 = mg\cdot 10m\\\\V = \sqrt[]{2g\cdot 10m} \\

And so we have found the velocity of the ball as it hits the floor.

V = \sqrt[]{2g\cdot 10m}=14.01\, m/s

Now, after the ball has bounced, we can again do an energy analysis, and we will get the same result, namely:

V = \sqrt[]{2g\cdot h}

where h is the maximum height of the ball, and v is the maximum speed of the ball (which is always attained at the bottom). If we know that now the height the ball achieves is 4 meters, plugging that in:

V = \sqrt[]{2g\cdot 4m} =8.86 \, m/s

Now for C, we need to know for how long the ball will be in the air from the time we drop it from 10 meters, and how long it will take the ball to reach its new maximum height of 4 meters.

As the acceleration of gravity is a constant, that means that the velocity of the ball will change at a constant rate. When something changes at a constant rate, what is its average?  It's the average between initial and final velocity, look at diagram to understand. The area under the Velocity vs time curve is the displacement of the ball, and:

V_{avg}\cdot t=h\\t=h/V_{avg}

what's the average speed when the ball is descending?

V_{avg}=\frac{1}{2} (14.01\, m/s+0)=7 \, m/s

so the time it takes the ball to go down is:

t=h/V_{avg}=\frac{10m}{7m/s} =1.43s\\

Now, when it goes up, it's final and initial speeds are 0 and 8.86 meters per second, thus the average speed is:

V_{avg}=\frac{1}{2} (8.86\, m/s+0)=4.43 \, m/s

and the time it takes to go up is:t=h/V_{avg}=\frac{4m}{4.43m/s} =0.90s

When we add both times , we get:

t_{total}=t_{down}+t_{up}=1.43s+0.90s = 2.33s

6 0
3 years ago
What in the atmosphere hold in the amount of heat needed to keep earth livable
aivan3 [116]

Answer:

Greenhouse gases keep our planet livable by holding onto some of Earth's heat energy so that it doesn't all escape into space. This heat trapping is known as the greenhouse effect. Just as too little greenhouse gas makes Earth too cold, too much greenhouse gas makes Earth too warm.

5 0
3 years ago
A heat pump absorbs heat from the atmosphere at a rate of 30 kW. If work is being done to run this heat pump at a rate of 7.7 kW
Volgvan

Answer:

Option D 3.9

Explanation:

First, you need to use the correct equation which is the following:

COP = Q/W

Where:

Q = heat absorbed

W = work done by the pump

COP = coefficient of perfomance

We have all the data, so, all you need to do is replace in the above expression and you shoould get the correct result:

COP = 30 / 7.7

COP = 3.896

This result you can round it to 3.9. option D.

8 0
3 years ago
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