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storchak [24]
2 years ago
15

Plss s ss s s s s s s s s

Physics
2 answers:
Crazy boy [7]2 years ago
8 0

Answer:

i font know ewan madaling sabihin yan

hichkok12 [17]2 years ago
7 0
Hey, I’ll may be wrong but here it is!
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There is a current of 0.83 A through a lightbulb in a 120 V circuit. What is the resistance of this lightbulb?
Lemur [1.5K]

Considering the Ohm's law, the resistance of the lightbulb is 144.58 Ω.

<h3>Definition of current</h3>

The flow of electricity through an object, such as a wire, is known as current (I). Its unit of measure is amps (A). So the current is a measure of the speed at which the charge passes a given reference point in a specified direction.

<h3>Definition of voltage</h3>

The driving force (electrical pressure) behind the flow of a current is known as voltage and is measured in volts (V) (voltage can also be referred to as the potential difference or electromotive force). That is, voltage is a measure of the work required to move a charge from one point to another.

<h3>Definition of resistance</h3>

Resistance (R) is the difficulty that a circuit opposes to the flow of a current and it is measured in ohms (Ω).

<h3>Ohm's law</h3>

Ohm's law establishes the relationship between current, voltage, and resistance in an electrical circuit.

This law establishes that the intensity of the current that passes through a circuit is directly proportional to the voltage of the same and inversely proportional to the resistance that it presents.

Mathematically, Ohm's law is expressed as:

I=\frac{V}{R}

Where:

  • I is the current measured in amps (A).
  • V the voltage measured in volts (V).
  • R the resistance that is measured in ohms (Ω).

<h3>Resistance of the lightbulb</h3>

In this case, you know that the voltage between two points in a circuit is 120 V and there is a current of 0.83 A.

Replacing in the Ohm's Law:

0.83A=\frac{120 V}{R}

Solving:

0.83 A× R=  120 V

R=\frac{120 V}{0.83A}

<u><em>R= 144.58 Ω</em></u>

Finally, the resistance of the lightbulb is 144.58 Ω.

Learn more about Ohm's law:

brainly.com/question/13076023

brainly.com/question/17286882

brainly.com/question/2275770

#SPJ1

3 0
1 year ago
6. Convert -180° to radians
icang [17]

Answer:

-3.141593

Explanation:

3 0
3 years ago
Read 2 more answers
Two pipes move the same amount of ideal fluid in the same amount of time. One pipe has a 2 in. diameter; the other has a 3 in. d
KATRIN_1 [288]

Answer:

a) 3-in. pipe

Explanation:

Given that

Fluid flow is in same amount in the same time it means that volume flow rate is same for the pipes

Volume flow rate

Q = A V

A=Area ,V=Velocity

A=\dfrac{\pi}{4}d^2

If diameter d is more then the velocity will be less for same volume flow rate .We also Know that if pressure is more then the velocity will be less.

The second pipe 3 in diameter having more diameter then the velocity will be less but the pressure will be more.

That is why the 3 in diameter is having more pressure than 2 in diameter pipe.

Therefore the answer will be a.

a) 3-in diameter  pipe

6 0
3 years ago
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th
Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

E_i = E_f

0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

PART B) Replacing the values given as,

h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\

v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}

v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}

v_f = 1.8486m/s

Therefore the speed of the masses would be 1.8486m/s

6 0
3 years ago
Aristotle supported which of these views?
katrin2010 [14]

The answer is A. The Sun and all the planets revolve around Earth.

Aristotle believed that the Earth was the centre of the solar system, and the Sun and the planets orbited around it. He believed that the universe was composed of Earth-like bodies, which were at rest, and of heavenly bodies, which were in perpetual motion.

3 0
2 years ago
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