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Vsevolod [243]
2 years ago
3

In a circuit consisting of two lamps connected in parallel, if there is 6 V across one lamp, what is the voltage across the othe

r lamp?
Please help!
Physics
1 answer:
Tatiana [17]2 years ago
7 0
If the two lamps are connected in parallel, then there is nothing
but wire between both of their top terminals, and nothing but wire
between their bottom terminals.

The result is that their bottom terminals are connected together,
their top terminals are connected together, the voltages at
corresponding points of both lamps are exactly the same, and
the total voltages ACROSS both lamps are equal.

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If you double the mass of an object while leaving the net force unchanged what is the result
valentinak56 [21]

Answer: If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

Explanation:

5 0
2 years ago
Two electric force vectors act on a particle. Their x-components are 13.5 N and −7.40 N and their y-components are −12.0 N and −
guapka [62]

Answer:

Explanation:

Given two vectors as follows

E₁ = 13.5 i -12 j

E₂ = -7.4 i - 4.7 j

Resultant E = E₁ + E₂

= 13.5 i -12 j -7.4 i - 4.7 j

E = 6.1 i - 16.7 j

a ) X component of resultant = 6.1 N

b ) y component of resultant = -16.7 N

Magnitude of resultant = √ ( 6.1² + 16.7² )

= 17.75 N

d ) If θ be the required angle

tanθ = 16.7 / 6.1 = 2.73

θ = 70° .

counterclockwise = 360 - 70 = 290°

6 0
2 years ago
Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th
puteri [66]
Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.
6 0
2 years ago
Read 2 more answers
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
A sister spins her brother in a circle of radius R at angular speed wi. Then the sister decreases her angular speed
Igoryamba

Answer:

The change in the centripetal acceleration of the brother,

                               Δa = V₂²/R - V₁²/R

Explanation:

Given data,

A sister spins her brother in a circle of radius, R

The angular velocity of the brother, ω₁ = V₁/R

The angular velocity of the brother, ω₂ = V₂/R

The centripetal acceleration is given by the relation

                                 a = V²/R

Therefore change in the centripetal acceleration of the brother,

                              Δa = V₂²/R - V₁²/R                                    

6 0
2 years ago
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