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dexar [7]
3 years ago
13

If 2.0 x 10^-4 C charge passes a point in 5.0 x 10^-5 s, what is the rate of current flow?

Physics
2 answers:
vampirchik [111]3 years ago
8 0
Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:

I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>
VladimirAG [237]3 years ago
6 0

Q= I×t

Where Q is charge in Coulombs C,

I is current in Amperes A,

t is in seconds s.

The rate of current flow is defined as the charge passing through a point in the circuit per second.

Rearranging formula gives I=\frac{Q}{t}

⇒I = \frac{2.0*10^-4}{5.0*10^-5} = 4 Amperes

∴ current flow = 4.0×10^0 A

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The model of the atom proposed by Greek philosophers appears similar to the model proposed centuries later by Dalton. What was t
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3 years ago
The sound from a bolt of lightning travelled 4.08 km in 12.0 s. What was the speed
LenaWriter [7]

Answer:

83.67 m/s

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Set up a calculation to convert units of measure to what you need.

You have km/s and you need m/s.

4.08km     1000 m         83.67m

-----------  X ----------  =  ---------------   the km will cancel out and you are left

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6 0
2 years ago
The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie
Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

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

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I'  and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0
2 years ago
Read 2 more answers
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