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Schach [20]
3 years ago
8

1-Calculate Req

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
3 0

Answer:

1. 21.66 Ohms

2.  3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

  Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

  Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms

2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps.  I am not an expert in physics but its ok :)

<u><em>Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.</em></u>

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What is the magnification when an object is placed at 2f from the pole of the convex mirror? 
Gekata [30.6K]

Answer:

Linear magnification = 1/3

Explanation:

Given:

Convex mirror

Object's distance from pole = 2f

Find:

Linear magnification

Computation:

Object distance, u = −2f

So,

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/f

1/v = 1/f + 1/2f

BY taking LCM

1/v = 3 / 2f

v = 2f / 3

Magnification, M = -v / u

So,

Magnification, M = (2f / 3) / 2f

Magnification, M = 2f / 6f

Magnification, M = 2 / 6

Linear magnification = 1/3

 

5 0
2 years ago
The average distance from Earth to the Moon is 384,000 km. In the late 1960s, astronauts reached the Moon in about 3 days. How f
Anon25 [30]

Answer:

They must have been traveling at 5333.33 km/h to cover that distance in 3 days.

That speed are 6,66 times higher than the speed of an aircraft jet.

Explanation:

d= 384000 km

t= 3 days = 3*24hr = 72hr

V= 384000km/72hr

V= 5333.33 km/h

comparison:

V1/V2= 5333.33/800

V1/V2= 6.66

3 0
3 years ago
A 12 V battery is connected to a 1200 Ω resistor. How much current is flowing through the resistor?
Oduvanchick [21]

Explanation:

Use Ohm's law.

V = IR

12 V = I (1200 Ω)

I = 0.01 A

7 0
3 years ago
A strong lightning bolt transfers an electric charge of about 31 C to Earth (or vice versa). How many electrons are transferred?
olchik [2.2K]

Answer:

m=5.78\times 10^{-3}\ g

n_e=1.935\times 10^{20} is the no. of electrons

Explanation:

Given:

  • quantity of charge transferred, Q=31\ C

<u>No. of electrons in the given amount of charge:</u>

As we have charge on one electron 1.602\times 10^{-19}\ C

so,

n_e=\frac{Q}{e}

n_e=\frac{31}{1.602\times 10^{-19}}

n_e=1.935\times 10^{20} is the no. of electrons

  • Now if each water molecules donates one electron:

Then we require n=1.935\times 10^{20} molecules.

<u>Now the no. of moles in this many molecules:</u>

n_m=\frac{n}{N_A}

where

N_A=6.022\times 10^{23} Avogadro No.

n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}

n_m=3.213\times 10^{-4}\ moles

  • We have molecular mass of water as M=18 g/mol.

<u>So, the mass of water in the obtained moles:</u>

n_m=\frac{m}{M}

where:

m = mass in gram

3.213\times 10^{-4}=\frac{m}{18}

m=5.78\times 10^{-3}\ g

7 0
3 years ago
The milky way is often considered to be an intermediately wound, barred spiral, which would be type ________ according to hubble
lora16 [44]
SBb type spiral. Its a type B because its not too tightly wound but its still too tight to be a type C
4 0
3 years ago
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