Ask question

Ask question

All categories

2 years ago

11

Lamp power with what is the current through the lamp when it is connected to a battery ?

1 answer:

shtirl [24]2 years ago

4 0

Answer:

lamp power with what is the current through the lamp when it is connected to a battery ?

Explanation:

If the light bulbs are connected in parallel, the current flowing through the light bulbs combine to form the current flowing in the battery, while the voltage drop is 6.0 V across each bulb and they all glow. One bulb burning out in a series circuit breaks the circuit.

You might be interested in

A(n) 82.7 kg boxer has his first match in the Canal Zone with gravitational acceleration 9.782 m/s 2 and his second match at the

Annette [7]

Answer:

82.7 kg

Explanation:

the mass of the boxer remains unchanged, this is because mass is a measure of the quantity of matter in an object irrespective of its location and the gravitational force acting at its location. this means mass is independent of the gravitational acceleration hence it remains the same 82.7 kg. its unit is in kilograms (Kg).

6 0

2 years ago

Will give correct answer brainliest

lukranit [14]

Answer:

I THINK it’s A

Explanation:

Because all the other answers don’t make sense.

5 0

2 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

I = 220 A

7 0

2 years ago

Consider the speed of light in another universe to be only 100 m/s. Two cars are traveling along a highway in opposite direction

olganol [36]

Answer:

-62.45m/s and +62.45m/s

Explanation:

The formula for relativistic speed

This is the speed of A with respect to B

$V_{AB}=\frac{V_{A}-V_{B}}{1-\frac{V_{A}V_{B}}{C^2} }$

where

$V_{A}$ will be the velocity of person 1: 39m/s

$V_{B}$ will be the velocity of person 2: -31m/s (negative because is travelling in opposite direction)

and $C$ the velocity of light: 100m/s

The velocity of person 1 measured by person 2 is:

$V_{AB}=\frac{39m/s-(-31m/s)}{1-\frac{(39m/s)(-31m/s)}{(100m/s)^2}}=62.45m/s$

The velocity of person 2 measured by person 1 is:

$V_{BA}=\frac{V_{B}-V_{A}}{1-\frac{V_{B}V_{A}}{C^2} }$

$V_{BA}=\frac{-31m/s-39m/s}{1-\frac{(-31m/s)(39m/s)}{(100)^2} }=-62.45m/s$

8 0

2 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

7 0

3 years ago