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3 years ago

Gawaingnpang nkabuhayan,hamon at oportinidad

Physics

1 answer:

Savatey [412]3 years ago

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gawaingnpang nkabuhayan, hamon at oportinidad

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Solving a series circuit, did I do this correctly?

nirvana33 [79]

The total resistance in the circuit is 16 Ohms.
The total current in the circuit is 0.5 Ampere.
The current at $R_1$ is 0.5 Ampere.
The current at $R_3$ is 0.5 Ampere.
The voltage drop at $R_1$ is 4 volts.
The voltage drop at $R_2$ is 2.5 volts.
The voltage drop at $R_3$ is 1.5 volts.
The total power consumed by the circuit is 4watts
The power consumed at $R_1$ is 2 watts
The power consumed at $R_2$ is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

$R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega$

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

The total resistance in the circuit = $R_{eq}$

$R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega$

The total current in the circuit = I

$V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A$ (Ohm's law)

For series combinations, the current in each resistor remains the same.

So, the current in $R_1, R_2 \&R_3$ :

$I_1= I_2= I_3=I=0.5 A\\$

The voltage drop across at $R_1$ = $V_1$

The current across $R_1$ = I = 0.5 A

$V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V$

The voltage drop across at $R_2$ = $V_2$

The current across $R_2$ = I = 0.5 A

$V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V$

The voltage drop across at $R_3$ = $V_3$

The current across $R_3$ = I = 0.5 A

$V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V$

The total power consumed by circuit:

$P= V\times I \\\\= 0.5 A\times 8 V = 4 watt$

Power consumed at $R_1$ :

$P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt$

Power consumed at $R_2$ :

$P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt$

Power consumed at $R_3$ :

$P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt$

Learn more about, current, voltage, resistance, and power of the circuit here:

brainly.com/question/11683246?referrer=searchResults

brainly.com/question/1430450?referrer=searchResults

5 0

2 years ago

Name at least two things that aren’t inherited

Natali5045456 [20]

1. Life
2. Your grades at school

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3 years ago

Read 2 more answers

A moving particle is subject to conservative forces only. When its kinetic energy decreases by 10 J, what happens to its mechani

satela [25.4K]

Answer:

Its mechanical energy is the same.

Explanation:

If forces are only conservative, the mechanical energy will be the same.

It can be different if energy get transformed in another kind of energy like elastic energy for example, although the amount of energy is always the same.

If we just have mechanical energy not geting transformed we have:

Em=K+U

Em: Mechanical energy

K: Kinetic energý

U: Potential energy

Then if Kinetic energy decreases 10J, Potential energy will grow up 10J to keep the same amount of mechanical energy.

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3 years ago

Choose the law each sentence describes. This law relates a planet's orbital period and its average distance to the Sun. The orbi

hram777 [196]

These are the Kepler's laws of planetary motion.

This law relates a planet's orbital period and its average distance to the Sun. - Third law of Kepler.

The orbits of planets are ellipses with the Sun at one focus. - First law of Kepler.

The speed of a planet varies, such that a planet sweeps out an equal area in equal time frames. - Second law of Kepler.

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3 years ago

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Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket's engine is to be fired

ValentinkaMS [17]

Answer:

$\frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=1096.633$

Explanation:

$m_i$ = Initial mass of rocket

$m_f$ = Final mass of rocket

u = Initial velocity

$v_r$ = Relative velocity

v = Velocity

From the rocket equation

$v=u+v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v}}=e^1\\\Rightarrow \frac{m_i}{m_f}=2.7182$

$\frac{m_i}{m_f}=2.7182$

when $v=7v_r$

$v=v_{r}\ln {\frac {m_{i}}{m_{f}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{v}{v_{r}}}\\\Rightarrow \frac{m_i}{m_f}=e^{\frac{7v_r}{v_r}}=e^7\\\Rightarrow \frac{m_i}{m_f}=1096.633$

$\frac{m_i}{m_f}=1096.633$

6 0

3 years ago

Gawaingnpang nkabuhayan,hamon at oportinidad​

Gawaingnpang nkabuhayan,hamon at oportinidad