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Korolek [52]
4 years ago
14

Please help! Picture Attached

Physics
2 answers:
Aleksandr [31]4 years ago
5 0
HEY THERE!!

ANSWER:- OPTION ( D )

HOPE IT HELPS YOU.
jeka944 years ago
4 0
I think it’s D
I hope this hopes
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20 points plz help i dont really understand how to do this
Artist 52 [7]

\huge{ \mathfrak{  \underline{ Answer} \:  \:  ✓ }}

1. Calculate the equivalent resistance in the circuit :

In the given diagram, the resistors are in series connection,

So equivalent resistance = sum of resistance of both resistors.

\large \boxed{R_{eq} =R_1 +R_2  }

  • R_{eq} =  480 + 360

  • R_{eq} = 840 \: ohms

Therefore, Equivalent resistance = 840 ohms.

2. Calculate the current through the battery :

\boxed{ \mathrm{current =  \dfrac{potential \:  \: difference}{resistance} }}

  • i =  \dfrac{120}{840}

  • i =  \dfrac{1}{7}

  • i = 0.142 \: Amperes

Hence, current through the battery = 0.142 A

3. How much current is passing through each resistor :

Since the resistors are joined in series connection, the current flowing through each resistor will be equal = 0.142 Amperes.

___________________________

\mathrm{ \#TeeNForeveR}

4 0
3 years ago
What is the function of a neutral operant?
zlopas [31]
B because it is correct
7 0
3 years ago
Read 2 more answers
What is the potential engery of a 150 kg diver standing on a diving board that is 10m high?
krek1111 [17]
The Answer Is Going To Be14,700
5 0
3 years ago
How do you complete this circuit?
ale4655 [162]

Explanation:

jejjwnwjejjejejejejrhnrn noe

7 0
4 years ago
The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,
omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

g = \frac{GM}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

M_p = \frac{1}{100} M_e

r_p = \frac{1}{4} r_e

The relationship of gravity would then be given:

g_e = \frac{GM_e}{r_e^2}

The relationship with the new planet, from the gravity of the earth would be given

g_p = \frac{GM_p}{r_p^2}

g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}

g_p = \frac{GM_e 16}{100 r_e^2}

g_p = 0.16 \frac{GM_e}{r_e^2}

g_p = 0.16g_e

The relationship with the weight of the earth would be given as:

W_e = m*g_e = 600N

W_p = m*g_p = m(0.16g_p)

W_p = (m*g_p)(0.16)

W_p = 600*0.16

W_p = 96N

Therefore the weigh on this planet would be 96N

3 0
3 years ago
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