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marusya05 [52]
3 years ago
15

Who do the basic particles of a compound differ from the basic particles of an element

Physics
1 answer:
Musya8 [376]3 years ago
4 0
A compound is elements bonding together, an element is the pure substance
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A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart
serious [3.7K]

To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

KE_{final} = (PE_{initial}+KE_{initial} )-PE_{final}

KE_{final} = (5100+4200)-5700

KE_{final} = 3600MJ

Therefore the final kinetic energy is 3600MJ

5 0
3 years ago
Will mark as brainliest if correct!!!!!!!
lutik1710 [3]

C. A could be ruby (speed of light = 170,000 km/s); B could be diamond (speed of light = 120,000 km/s).

Explanation:

Refraction is a phenomenon that occurs when a light rays crosses the boundary between two different mediums.

When this occurs, the light wave changes speed and also direction, according to Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2 (1)

where

n_1 is the index of refraction of the first medium

n_2 is the index of refraction of the second medium

\theta_1 is the angle of incidence (the angle between the incident ray and the normal to the boundary)

\theta_2 is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

The index of refraction is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):

n=\frac{c}{v}

Using this definition, we can rewrite eq.(1) as

\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}

Where v_2 is the speed of light in the 2nd medium and v_1 the speed of light in the 1st medium.

Now let's analyze the situation represented in the figure: we see that as the light ray enters the 2nd medium, it bends towards the normal, this means that the angle of refraction is smaller than the angle of incidence:

\theta_2 < \theta_1

This means that

\frac{\sin \theta_2}{\sin \theta_1}

And therefore,

\frac{v_2}{v_1}

So, the speed of light in the second medium is smaller than the speed of light in the first medium: this occurs only in option C), which is therefore the correct choice.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

6 0
3 years ago
Help asap pls. Which of the following best describes how high and low pitch are different?
Gwar [14]
Answer D. The pitch is in the frequency.
3 0
2 years ago
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb
kramer

Answer:

12552 J or 3000 calories

Explanation:

Q = m × c × ∆T

Where;

Q = amount of heat energy (J)

m = mass of water (g)

c = specific heat capacity (4.184 J/g°C)

∆T = change in temperature

For 50mL of water, there are 50g, hence, m = 50g, c = 4.184 J/g°C, initial temperature = 0°C, final temperature = 60°C.

Q = m × c × ∆T

Q = 50 × 4.184 × (60 - 0)

Q = 209.2 × 60

Q = 12552 J

Hence, the amount of heat energy used to heat the water is 12552 J or 3000 calories

3 0
3 years ago
Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. Ho
lidiya [134]

Answer:\frac{r_1}{r_2}=1.565

Explanation:

Given

two holes are made with different sizes

Hole 1 is large in size and hole 2 is small

If the volume flow rate of water is same for both the hole then small hole must be below the large hole because for same flow rate, velocity of water is large while cross-sectional area is small so it compensate to give same flow for both the holes.

Now for radius apply Bernoulli's theorem at hole 1 and 2

P_1+\rho gh_1=P_{atm}+\frac{1}{2}\rho v_1^2

P_2+\rho g6h_2=P_{atm}+\frac{1}{2}\rho v_2^2

if hole 1 is h distance below water surface then h_2=6h

and P_1=P_2=P_{atm}

Also v_1=\sqrt{2gh}

v_2=\sqrt{2g(6h)}

and Q=A_1v_1=A_2v_2

A=\pi r^2

thus \dfrac{r_1}{r_2}=\sqrt{\dfrac{v_2}{v_1}}

\dfrac{r_1}{r_2}=\sqrt{\dfrac{\sqrt{6h}}{\sqrt{h}}}

\frac{r_1}{r_2}=1.565

5 0
3 years ago
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