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salantis [7]
3 years ago
14

As a space shuttle moves through the dilute ionized gas of Earth's ionosphere, the shuttle's potential is typically changed by -

1.4 V during one revolution. Assuming the shuttle is a conducting sphere of radius 15 m, estimate the amount of charge it collects.
Physics
1 answer:
postnew [5]3 years ago
4 0

Answer:

-2.3 × 10^-9 Coulombs(C).

Explanation:

So, we are given the following data or information or parameters that is going to help us to solve the problem effectively and efficiently;

=> " the shuttle's potential is typically changed by -1.4 V during one revolution. "

=> " Assuming the shuttle is a conducting sphere of radius 15 m".

So, in order to estimate the value for the charge we will be making use of the equation below:

Charge, C =( radius × voltage or potential difference) ÷ Coulomb's law constant.

Note that the value of Coulomb's law constant = 9 x 10^9 Nm^2 / C^2.

So, charge = { 15 × (- 1.4)} / 9 x 10^9 Nm^2 / C^2.

= -2.3 × 10^-9 Coulombs(C).

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3 years ago
When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.3 a while
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Correct question: When two resistors are connected in parallel across a battery of unknown voltage, one resistor carries a current of 3.3 A while the second carries a current of 1.8 A .

What current will be supplied by the same battery if these two resistors are connected to it in series?

Answer:

1.164 A

Explanation:

From the question,

Let the unknown voltage of the batter = V.

Since the two resistors are connected in parallel,

The voltage across each of the resistor is the same = V,

Then,

From ohm's law,

V = IR................... Equation 1

Where V = Voltage, I = Current, R = Resistance.

make R the subject of the equation

R = V/I................ Equation 2

For the first resistor,

Given: I₁ = 3.3 A, V = V.

Substitute into equation 2

R₁ = V/3.3 Ω

For the second resistor,

Given: I₂ = 1.8 A

Substitute into equation 2

R₂ = V/1.8 Ω.

When the Resistors are connected in series,

Rt = R₁+R₂

Where Rt = Total resistance.

Rt = V/3.3+V/1.8

Rt = (1.8V+3.3V)/(3.3×1.8)

Rt = 5.1V/5.94

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Applying,

I' = V/Rt

where I' = current supplied by the battery, If the two resistors are connected in series

I' = V/0.859V

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