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3 years ago

Why at the poles is the cenripetal accleration zero?

1 answer:

7 0

The poles are essentially a fixed point.

I have so much to say but won't be able to articulate it in a succinct manner, sorry for keeping it short and sweet.

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In order to be considered an ion, an atom must have a

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Answer is B, it can be positive or negative, as long as it has a charge.

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3 years ago

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Which of the following is an example of qualitative data?

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10 grams of salt is the result

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3 years ago

What is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k,

WITCHER [35]

The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force. This is the significance of the fact that the gravitational constant, G, is a very small number and that Coulomb’s constant, k, is a very large number

Answer: The comparison indicates that the gravitational force is a much weaker force than the electromagnetic force.

Explanation:

We know that the universal formula for the estimation of gravitational force is,

$F=G \times\left(\frac{m_{1} \times m_{2}}{r^{2}}\right)$

Where,

F, Gravitational force, $m_{1} \text { and } m_{2}$ - objects masses r- The relative distance between the two objects

G, Gravitational Constant $6.67 \times 10^{-11} \frac{N m^{2}}{k g^{2}}$

And, the electrostatic force between two scalar charges according to the Coulomb’s law is,

$F=k \times\left(\frac{q_{1} \times q_{2}}{r^{2}}\right)$

F, Electrostatic force between two charges, $q_{1} \text { and } q_{2}$ - Two scalar charges, r - The relative distance between two scalar charges, k- Coulomb’s constant $8.987 \times 10^{9} \frac{N m^{2}}{c^{2}}$

Now, on comparing the values of G and k, we can easily evaluate that eventually, the gravitational force will be lesser than the coulomb’s force.

Besides this, we can also judge this fact through various examples such as, a balloon rubbed with a cloth, easily sticks to the wall for some time. Opposing the gravitational force of the Earth which is not the case with the normal balloon.

It drops without having the electrostatic force between the wall and the balloon. This shows that the gravitational force draws lesser impact on objects as compared to the electrostatic force.

7 0

3 years ago

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Find A and effective resistance. (Ill give Brainliest if you provide explaination)

Alex777 [14]

Answer:

A = 2.4 A

$R_{eq} = 5 \ \Omega$

Explanation:

The voltage in the circuit, V = 12 V

The given circuit shows four resistors with R₁ and R₂ arranged in series with both in parallel to R₃ and R₄ which are is series to each other

R₁ = 4 Ω

R₂ = 6 Ω

R₄ = 5 Ω

The voltage across R₃ = 6 V

Voltage across parallel resistors are equal, therefore;

The total voltage across R₃ and R₄ = 12 V

The total voltage across R₁ and R₂ = 12 V

The voltage across R₃ + The voltage across R₄ = 12 V

∴ The voltage across R₄ = 12 V - 6 V = 6 V

The current flowing through R₄ = 6V/(5 Ω) = 1.2 A

The current flowing through R₃ = The current flowing through R₄ = 1.2 A

The resistor, R₃ = 6 V/1.2 A = 5 Ω

Therefore, we have;

The sum of resistors in series are R₁ + R₂ and R₃ + R₄, which gives;

$R_{series \, 1}$ = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω

$R_{series \, 2}$ = R₃ + R₄ = 5 Ω + 5 Ω = 10 Ω

The sum of the resistors in parallel is given as follows;

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_{series \, 1}} + \dfrac{1}{R_{series \, 2}} = \dfrac{R_{series \, 2} + R_{series \, 1}}{R_{series \, 1} \times R_{series \, 2}}$