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Harrizon [31]
3 years ago
11

A source charge of 3 µC generates an electric field of 2.86 × 105 N/C at the location of a test charge. Using k = 8.99 × 109N.m^

2/c^2, what is the distance, to the
Physics
2 answers:
viktelen [127]3 years ago
9 0

0.31 for E2020 users

Nataliya [291]3 years ago
7 0
Variables:

Source charge, Q = 3 micro C = 3 * 10^ - 6 C

E = electric field = 2.86 * 10 ^5 N/C

K = 8.99 * 10^9 N * m^2 / C

d = distance = ?

Formula:

E = K * Q / (d^2) => d^2 = K * Q / E

=> d^2 = 8.99 * 10^9 N * m^2 / C * 3 * 10^ -6 C / (2.86 * 10^ 5 N/C)

d^2 = 9.43 * 10 ^ -2  m^2

=> d = 3.07 * 10^-1 m

Answer: 0.307 m

Note: it is a long distance due to the Electric field is very low
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21. A person standing 49.5m from the foot of a cliff claps his hands and hears an echo 0.3
777dan777 [17]

Answer:

330 m/s

Explanation:

The sound wave has to travel TO the cliff AND back = 2 * 49.5 = 99 m

magnitude of velocity =  distance / time = 99m / .3 s = 330 m/s

5 0
1 year ago
An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr
Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

F= mg

Where, f = restoring force

kx=mg

k=\dfrac{mg}{x}

Put the value into the formula

k=\dfrac{2.15\times9.8}{0.0895}

k=235.41\ N/m

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

Here, v = A\omega

K.E=\dfrac{1}{2}m\times(A\omega)^2

Here, \omega=\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2

K.E=\dfrac{1}{2}kA^2

Put the value into the formula

K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2

K.E=0.06500\ J

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

8 0
3 years ago
Help me rearrange this formula. <br><br>I've been trying but I can't remember how to do it.​
NikAS [45]

-- Multiply each side of the formula by 2

-- Then divide each side by t

-- Then subtract V(i) from each side.

7 0
3 years ago
Can you please help me
konstantin123 [22]

Answer:

I dont know sorryyyyyyy

4 0
3 years ago
When astronomers look at distant galaxies, what sort of motion do they see?
arlik [135]
Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

Please hit that thank you button if that helped, I don’t want thank you’s I just want to know that this helped.

Please reply if this doesn’t help, I will try my best to gather more information or a answer.

Here is some good information that could help you out a lot!


Let’s begin by exploring some techniques astronomers use to study how galaxies are born and change over cosmic time. Suppose you wanted to understand how adult humans got to be the way they are. If you were very dedicated and patient, you could actually observe a sample of babies from birth, following them through childhood, adolescence, and into adulthood, and making basic measurements such as their heights, weights, and the proportional sizes of different parts of their bodies to understand how they change over time.

Unfortunately, we have no such possibility for understanding how galaxies grow and change over time: in a human lifetime—or even over the entire history of human civilization—individual galaxies change hardly at all. We need other tools than just patiently observing single galaxies in order to study and understand those long, slow changes.

We do, however, have one remarkable asset in studying galactic evolution. As we have seen, the universe itself is a kind of time machine that permits us to observe remote galaxies as they were long ago. For the closest galaxies, like the Andromeda galaxy, the time the light takes to reach us is on the order of a few hundred thousand to a few million years. Typically not much changes over times that short—individual stars in the galaxy may be born or die, but the overall structure and appearance of the galaxy will remain the same. But we have observed galaxies so far away that we are seeing them as they were when the light left them more than 10 billion years ago.


That is some information, I do have more if you need some! Thanks!

Have a great rest of your day/night! :)


Emacathy,
Brainly Team.


8 0
3 years ago
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