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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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<u>Answer</u>: The potential difference across the resistor is 12 volts.

<u>Explanation:</u>

To calculate the potential difference cross the resistor, we use Ohm's Law. This law states that the potential difference across two wires is directly proportional to the current flowing through that wire.

Mathematically,

V\propto I\\V=IR

Where,

V = potential difference = ?V

I = Current flowing = 1.2 A

R = Resistor = 10\Omega

Putting values in above equation, we get:

V=1.2\times 10=12V

Hence, the potential difference across the resistor is 12 volts

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If an organism didn't have DNA, it would be missing which sign of life ? a, b , c
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A

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Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

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Answer:

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In the context, if the nucleus is compared to the size of a nonpareil then its atom would be of the size of small size marble. An atom is bigger in size than that of nucleus as the nucleus is located inside the atoms.

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