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irinina [24]
4 years ago
7

3. The circuit below shows two outdoor lights, run by a battery. Four amps of

Physics
1 answer:
miv72 [106K]4 years ago
4 0

Answer:

The answer to your question is given below.

Explanation:

From the diagram given in the question, we can see that the bulbs are in parallel connection. This implies that different current will go through the bulb but the same voltage. But the resistance of each bulb is the same, therefore the same current current will go through them.

This can further be understood if we determine the current flowing through each bulb. This is illustrated below:

Voltage (V) = 30V

Resistant 1 (R1) = 15Ω

Current 1 (I1) =..?

V = I1R1

30 = I1 x 15

Divide both side by 15

I1 = 30/15

I1 = 2A.

Voltage (V) = 30V

Resistant 2 (R2) = 15Ω

Current 2 (I2) =..?

V = I2R2

30 = I2 x 15

Divide both side by 15

I2 = 30/15

I2 = 2A.

The overall current(I) in the circuit:

I = I1 + I2

I = 2 + 2 = 4A.

From the illustrations above, we can see that the same current will flow through each bulb as the will current split into two equal parts

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3 years ago
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you are driving at 18m/s down Lyndale​ avenue. A car backs out a driveway 25 meters in front of you. You continue at that speed
zimovet [89]

Hello!

For this, first let's calculate time of stop:

t = (V - Vi) / a

Replacing:

t = (0 m/s - 18 m/s) / -5,4 m/s^2

Resolving:

t = -18 m/s / -5,4 m/s^2

t = 3,33 s + 0,25 s = 3,58 s

Now lets calculate distance traveled, with formula:

d = Vi*t + (a*t^2)/2

Replacing:

d = 18 m/s * 3,58 s + (-5,4 m/s^2 * (3,58 s)^2) /2

Resolving:

d = 64,44 m + (-34,604 m)

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3 years ago
You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6
marta [7]

Answer:

(a) The current needed is 56.92 A

(b) The magnitude of the magnetic field inside the solenoid is 0.134 T

(c) The energy density inside the solenoid is 7.144 kJ/m³

Explanation:

Given;

energy stored in the magnetic field of solenoid, E = 8.1 J

number of turns of the solenoid, N = 620 turns

diameter of the solenoid, D = 6.6 cm = 0.066 m

radius of the solenoid, r = D/2 = 0.033 m

length of the solenoid, L = 33 cm = 0.33 m

Inductance of the solenoid is given as;

L= \frac{\mu_o N^2 A}{l}

where;

A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²

μ₀ is permeability of free space = 4π x 10⁻⁷ H/m

L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H

(A). How much current needed

Energy stored in magnetic field of solenoid is given as;

E = \frac{1}{2} LI^2\\\\

Where;

I is the current in the solenoid

E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A

(B) The magnitude of the magnetic field inside the solenoid

B = μ₀nI

where;

n is number of turns per unit length

B = μ₀(N/L)I

B = (4π x 10⁻⁷)(620/0.33)(56.92)

B = 0.134 T

(C) The energy density (energy/volume) inside the solenoid

U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3

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Answer:

you a b1tc h

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