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Scrat [10]
3 years ago
8

Solve each problem. Be sure to show your work and give a final answer with units rounded to the given number of significant figu

res.
1) Find the voltage on a circuit with a resistance of 12.5 Ω if it has a current of 2.35 A.

2) If a circuit with a 9.0 V battery has a current of 6.2 A, how much resistance is in the circuit?

3) A circuit has three resistors with values of 4.0 Ω, 8.0 Ω, and 12 Ω. What is the equivalent resistance if they are wired in series?

4) A circuit has three resistors with values of 4.0 Ω, 8.0 Ω, and 12 Ω. What is the equivalent resistance if they are wired in parallel?
Physics
1 answer:
Alex Ar [27]3 years ago
8 0

Answer:

1. the voltage will be 2.35×12.5 = 29.4V

2. the resistance would be 9.0/6.2= 1.45ohms

3. in series they will add up thus 4+8+12= 24ohms

4. in parallel it will be 2.18ohms

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Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction
kumpel [21]

Answer:

The  correct option is  H

Explanation:

From the question we are told that

    The index of refraction of  coating is  n_1

       The  index of refraction of material  is  n_2

   

Generally the condition for constructive for a thin film interference is mathematically represented

            2 *  t  = [ m  + \frac{1}{2}] \frac{\lambda}{n_1 }

Here  t represents the thickness

For minimum thickness  m =  0

So

           2 *  t  =0  + \frac{1}{2}\frac{\lambda}{n_1 }

=>        t  =\frac{\lambda}{4n_1 }

3 0
3 years ago
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00°C, the resistance of the carbon r
dusya [7]

Answer:

28 degree C

Explanation:

We are given that

T_1=4.00^{\circ}

R_1=217.7 \Ohm

R_2=215.1\Ohm

\alpha=-5.00\times 10^{-4}C^{-1}

We have to find the temperature on a spring day when resistance is 215.1 ohm.

We know that

\alpha(T_2-T_1)=\frac{R_2}{R_2}-1

Using the formula

-5.00\times 10^{-4}(T_2-4)=\frac{215.1}{217.7}-1

-5\times 10^{-4}(T_2-4)=0.988-1=-0.012

T_2-4=\frac{0.012}{5\times 10^{-4}}=24

T_2=24+4=28^{\circ}C

Hence, the temperature  on a spring day 28 degree C.

7 0
3 years ago
Does anyone know? Please help! Thank you!
Vedmedyk [2.9K]

B density will increase

7 0
3 years ago
Read 2 more answers
Honeybees acquire a charge while flying due to friction with the air. A 100 mg bee with a charge of +23 pC experiences an electr
Sedbober [7]

Answer:

(A) ratio of electric force to weight will be  23.469\times 10^{-10}

(b) Electric field will be E=4.26\times 10^{10}N/C

Explanation:

We have given mass of bee = 100 mg  = m=100\times 10^{-3}=0.1kg

Charge on bee q=23pC=23\times 10^{-12}C

Electric field E = 100 N/C

Weight of the bee W=mg=0.1\times 9.8=0.98N

Electric force on the bee F=qE=23\times 10^{-12}\times 100=23\times 10^{-10}N

So the ratio of electric force on the bee and weight is =\frac{F}{W}=\frac{23\times 10^{-10}}{0.98}=23.469\times 10^{-10}

(B) To hold the bee in air electric force must be equal to weight of bee

So mg=qE

0.1\times 9.8=23\times 10^{-12}E

E=4.26\times 10^{10}N/C

4 0
3 years ago
In the diagram, q1=+6.25 * 10^ -8 C. What is the potential difference when you go from point A to point B? Include the correct s
Nimfa-mama [501]

Answer:

Moving a unit "positive" test charge from A to B will result in a reduction in potential

V = K Q / R      potential at a point

V2 - V1 = K Q (1 / .4 - 1 / .15) = = k Q (.15 - .4) / .06 = -4.17 K Q

V2 - V1 = -4.17 * 9 & 10E9 * 6.25 E-8

V2 - V1 = -4.17 * 562.5 J/C

V = - 2346 Volts

7 0
3 years ago
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