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poizon [28]
1 year ago
10

A 3. 5-a current is maintained in a simple circuit with a total resistance of 1500 ω. What net charge passes through any point i

n the circuit during a thirty second interval?.
Physics
1 answer:
melamori03 [73]1 year ago
7 0

Electric Current:

Electric current is the flow of charge through a given circuit per unit time. Electric current is one of the components needed to calculate the electric power that a device needs to operate and do work. Electric current is measured in amperes (A), which is equal to:

            1A = 1 C/ s

Recall that the coulomb (C) is the unit for charge while the second (s) is the unit for time

Given: I = 3.5

A is the current

Δt =30 s is the time interval

                     A =ΔQ/ΔT

Net charge = 100C

Electricity is produced when an electric current runs through a circuit.

How does electric current work?

A current of electricity is a steady flow of electrons. When electrons move from one place to another, round a circuit, they carry electrical energy from place to place like marching ants carrying leaves. Instead of carrying leaves, electrons carry a tiny amount of electric charge.

Learn more about Electric current :

brainly.com/question/27003377

#SPJ4

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A wire loop with 70 turns is formed into a square with sides of length ???? . The loop is in the presence of a 1.20 T uniform ma
Oksi-84 [34.3K]

Answer:

l= 3.002 cm

Explanation:

Given that

n= 70 turns

B= 1.2 T

θ= 15°

I= 1.5 A

τ = 0.0294 N⋅m

Lets take length of sides is l.

We know that

τ  = n I A B sin θ

Area of square ,A= l²

Now by putting the value

τ  = n I A B sin θ

0.0294  = 70 x  1.5 x  l² x  1.2 x  sin 15°

l² = 0.000901 m²

l² = 9.01 cm²

l= 3.002 cm

3 0
3 years ago
A fixed mass of gas of volume 600cm^3 at a temperature of 30°c is cooled to 0°c. At the same time, the volume is increased by 10
Lapatulllka [165]
Use Math-way it is very easy
7 0
2 years ago
14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?
Vladimir79 [104]
The relationship between resistance R and resistivity \rho is
R= \frac{\rho L}{A}
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m
and the cross section is
A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2

From the first equation, we can then find the length of the wire when R=4.8 \Omega (copper resistivity: \rho = 1.724 \cdot 10^{-8} \Omega m)
L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m
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7 0
3 years ago
A sheet of 8.5 inch by 11inch paper has a mass of 4.50 grams one square centimeter , 1cm2 of that sheet of paper will have a mas
finlep [7]
The answer is option D.

Area of sheet, A1  = length × breadth
                       =  (8.5 inch * 2.54 cm) ×11 inch * 2.54 cm) 
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Now, Mass of sheet, M1 = 4.50 gms

Thus, Area of sheet A2, = 1cm^{2}
And, Mass of sheet, M2 = ?

Thus,  Mass of M2 = \frac{4.50 gms}{603.22}
                                = 0.0073
3 0
3 years ago
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