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katovenus [111]
2 years ago
15

A 10.0 Ω lightbulb is connected to a 12.0 V battery. (a) What current flows through the bulb? (b) What is the power of the bulb?

Physics
1 answer:
andreev551 [17]2 years ago
6 0

The current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

<h3>What current flows through the bulb as well as the power of the bulb?</h3>

From ohm's law; V = I × R

Where V is the voltage, I is the current and R is the resistance.

Also, Power is expressed as; P = V × I

Where V is voltage and I is current.

Given that;

  • Resistance R = 10.0 ohms
  • Voltage V = 12.0V
  • Current I = ?
  • Power P = ?

First, we determine the current flow through the bulb.

V = I × R

12.0V = I × 10.0 ohms

I = 12.0 ÷ 10.0

I = 1.2A

Next, we determine the power of the bulb.

P = V × I

P = 12.0V × 1.2A

P = 14.4 Watts

Therefore, the current flowing through the bulb as well the power of the bulb are 1.2A and 14.4 Watts respectively.

Learn more about Ohm's law here: brainly.com/question/12948166

#SPJ1

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At a particular instant, a moving body has a kinetic energy of 295 J and a momentum of magnitude 25.1 kg · m/s.(a)What is the sp
motikmotik

Answer:

a) 23.51 m/s

b) 1.07 kg

Explanation:

Parameters given:

Kinetic energy, K = 295 J

Momentum, p = 25.1 kgm/s

a) The kinetic energy of a body is given as:

K = \frac{1}{2} mv^2

where m = mass of the body and v = speed of the body

We know that momentum is given as:

p = mv

Therefore:

K = 1/2 * pv

=> v = 2K / p

v = (2 * 295) / 25.1 = 23.51 m/s

The velocity of the body at that instant is 23.51 m/s.

b) Momentum is given as:

p = mv

=> m = p / v

m = 25.1 / 23.51  = 1.07 kg

The mass of the body at that instant is 1.07 kg

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Answer:

Explanation:

Force between two charges of q₁ and q₂ at distance d is given by the expression

F = k q₁ q₂ / d₂

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k = 1/ 4π x 8.85 x 10⁻¹²

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

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force between charge q₂ =  34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .

putting the values in the expression

F = 1/ 4π x 8.85 x 10⁻¹²  x  34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 9 x 10⁹ x  - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²

= 82729.6  x 10⁻³ N

Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)

Total force = 118699 x 10⁻³

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