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Alex73 [517]
3 years ago
5

a charge of 30. coulombs passes through a 24-ohm resistor in 6.0 seconds. what is the current through the resistor? (1) 1.3 a (3

) 7.5 a (2) 5.0 a (4) 4.0 a
Physics
2 answers:
DochEvi [55]3 years ago
7 0
As I = Q/t (charge/time) the answer is 5a.

30/6 = 5
uranmaximum [27]3 years ago
3 0
Current is the rate of flow of charges.

I = Q / t

Where I is current in Ampere,  Q is charges in Coulombs, t is time in seconds.

I = 30 / 6

I = 5 A.

Option (2) is the answer.
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Find the coefficient of kinetic friction μk. express your answer in terms of some or all of the variables d1, d2, and θ.
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6 0
3 years ago
Read 2 more answers
What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?
Taya2010 [7]

Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:

Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?

Answer:

The Force act on each side is 2.43×10⁴N

Explanation:

Given data

n=3 mol

L=0.3 m

Temperature=20.0°C=293 K

To find

Force F

Solution

To get force act on each side it would employ by

F=P.A

Where P is pressure

A is Area

First we need to find pressure by applying ideal gas law

So

P.V=nRT\\P=\frac{nRT}{V}\\ P=\frac{(3mol)(8.315J/mol)(293K)}{(0.3m*0.3m*0.3m)}\\P=27.069*10^{4}Pa

So The Force is given as:

F=P.A\\F=(27.069*10^{4} )(0.3m*0.3m)\\F=2.43*10^{4}N

The Force act on each side is 2.43×10⁴N

3 0
3 years ago
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Vedmedyk [2.9K]

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A wheel rotates about a fixed axis with an initial angular velocity of 20 rad/s. During a 5.0-s interval, the angular velocity d
Nikitich [7]

Answer:

The angular displacement of the wheel is 45 radians

Explanation:

Given;

initial angular velocity, ω₀ = 20 rad/s

final angular velocity, ωf = 10 rad/s

time interval, t = 5

Angular acceleration is calculated as;

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|α| = 2 rad/s²

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\theta = \omega_0 \  + \ \frac{1}{2} \alpha t^2\\\\\theta = 20 \ + \ \frac{1}{2} *(2)*5^2\\\\\theta = 20 \ + 25\\\\ \theta = 45 \ radians

Therefore, the angular displacement of the wheel is 45 radians

8 0
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