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dangina [55]
3 years ago
13

A simple circuit consists of a light bulb connected to the terminals of a battery. A voltmeter shows a potential difference of 1

8.5 V across the light bulb, whereas an ammeter registers a current of 0.485 A in the circuit. Determine the resistance of the bulb.
Physics
1 answer:
larisa [96]3 years ago
6 0
<h2>Answer:</h2>

38.14Ω

<h2>Explanation:</h2>

Let's solve this question using Ohm's law which states that the current (I) flowing through a conductor is directly proportional to the potential difference or voltage (V) across it. Mathematically;

V =  I R       -------------------(i)

<em>Where</em>;

R is the constant of proportionality called resistance of the conductor and is measured in Ohms (Ω)

<em>From the question;</em>

V = 18.5V

I = 0.485A

<em>Substitute these  values into equation (i) as follows;</em>

18.5 = 0.485 x R

<em>Solve for R;</em>

R = 18.5 / 0.485

R = 38.14Ω

Therefore the resistance of the bulb is 38.14Ω

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Black_prince [1.1K]

Answer:

a)  R = ρ₀ L /π(r_b² - R_a²) , b)  ρ₀ = V / I    π (r_b² - R_a²) / L

Explanation:

a) The resistance of a material is given by

          R = ρ l / A

where ρ is the resistivity, l is the length and A is the area

the length is l = L and the resistivity is ρ = ρ₀

the area is the area of ​​the cylindrical shell

           A = π r_b² - π r_a²

           A = π (r_b² - r_a²)

we substitute

         R = ρ₀ L /π(r_b² - R_a²)

b) The potential difference is related to current and resistance by ohm's law

         V = i R

         

we subsist the expression of resistance

          V = I ρ₀ L /π (r_b² - R_a²)

           ρ₀ = V / I    π (r_b² - R_a²) / L

6 0
3 years ago
If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

-- In order for both of them to add up to zero, the momentum of the cannon must be (2 kg-m/s this way <==) .

Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

Speed of recoil of the cannon = <em>-- 0.4 m/s</em>

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3 years ago
What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make
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Answer:

The transverse wave

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Timmy drove 2/5 of a journey at an average speed of 20 mph.
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Read 2 more answers
A wave of infrared light has a speed of 6m/s and a wave length of 12 m. What is the frequency of this wave?
erastovalidia [21]
Frequency = speed / wavelength

(6 m/s) / (12 m) = 0.5 Hz.

That's not infrared light.
Infrared light waves move about 50 million times faster than that, and they're only about 0.00000007 as long as that.
6 0
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