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vladimir2022 [97]
3 years ago
13

I WILL MARK BRAINLIEST!!ASAP!!! Wet Lab - Coulomb's Law lab from edge!!

Physics
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

h

Explanation:

Coulomb's law, or Coulomb's inverse-square law, is an experimental law[1] of physics that quantifies the amount of force between two stationary, electrically charged particles. The electric force between charged bodies at rest is conventionally called electrostatic force or Coulomb force.[2] The law was first discovered in 1785 by French physicist Charles-Augustin de Coulomb, hence the name. Coulomb's law was essential to the development of the theory of electromagnetism, maybe even its starting point,[1] as it made it possible to discuss the quantity of electric charge in a meaningful way.[3]

The law states that the magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them,[4]

{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}{\displaystyle F=k_{\text{e}}{\frac {q_{1}q_{2}}{r^{2}}}}

Here, ke is Coulomb's constant (ke ≈ 8.988×109 N⋅m2⋅C−2),[1] q1 and q2 are the signed magnitudes of the charges, and the scalar r is the distance between the charges.

The force is along the straight line joining the two charges. If the charges have the same sign, the electrostatic force between them is repulsive; if they have different signs, the force between them is attractive.

Being an inverse-square law, the law is analogous to Isaac Newton's inverse-square law of universal gravitation, but gravitational forces are always attractive, while electrostatic forces can be attractive or repulsive.[2] Coulomb's law can be used to derive Gauss's law, and vice versa. In the case of a single stationary point charge, the two laws are equivalent, expressing the same physical law in different ways.[5] The law has been tested extensively, and observations have upheld the law on the scale from 10−16 m to 108 m.[5]

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Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m
Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}.....(I)

tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}....(II)

Put the value of \omega=0.628\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}

A=0.0198

From equation (II)

tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}

d=0.0023

Put the value of \omega=3.14\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}

A=0.0203

From equation (II)

tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}

d=0.0120

Put the value of \omega=6.28\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}

A=0.0209

From equation (II)

tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}

d=0.0257

Put the value of \omega=9.42\ rad/s in equation (I) and (II)

A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}

A=0.0217

From equation (II)

tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}

d=0.0413

Hence, This is the required solution.

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3 years ago
Які лампочки у схемі з'єднані<br> паралельно?
alekssr [168]

Answer:

I don't understand please interpret

5 0
3 years ago
In which test did the air parcel rise the highest? Is there a pattern in the relationship between starting air temperature and p
wel

Answer:

xitxitx

Explanation:

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5 0
3 years ago
The closer the charged particles, the... the electrostatic force.
Anna35 [415]

Answer:

please give me brain list and follow

Explanation:

Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value. If the charges come 10 times closer, the size of the force increases by a factor of 100. The size of the force is proportional to the value of each charge.

4 0
3 years ago
Need help! Photo says all! Will mark brainliest :)
AVprozaik [17]

Answer:

C-D

Explanation:

As you can see from the graph, the distance from A to B was from 0 m to 6 m in a duration of 3 seconds.

Divide 6 meters by 3 seconds to find the speed:

6 ÷ 3 = 2 m/s

B-C is not moving due to a straight line as said in the graph, so speed is

0 m/s.

There is also C-D since the car traveled from a distance of 9 meters

(6 -(-3) = 9) in 3 seconds too. (NOTE: The graph line going down does not mean it is slowing down, but rather going to a certain distance like going backwards)

Divide 9 meters by 3 seconds to get the speed:

9 ÷ 3 = 3 m/s

Between A-B, B-C, and C-D, C-D has the fastest speed recorded with 3 m/s.

A-D does not count here as the line has no connection between point A and point D.

Cheers!

6 0
3 years ago
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