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3 years ago

Define resistance and describe what would happen to a lightbulb if the voltage increase but the resistance stay the same?

1 answer:

7 0

The first question is how much of a voltage increase are we looking at. If it has a 110 voltage rating and you put it across a 220 source, you will see one flash and then the bulb is no more. Nothing will revive it.

If it is rated at 110 and you put 130 across it, there's no problem but the bulb will burn out sooner than it would if you just put 110 across it.

So you raise the voltage and the resistance stays the same, the current will increase. That's why it will burn out sooner.

V = I * R

The equation is a direct variation. If the voltage goes up the current goes up. If the voltage goes down, the current goes down providing that the resistance stays the same in both cases.

The second question is what is resistance? Resistance in Electricity is the ability of an electric current to go in one direction freeing up as many electrons as it can. The MORE free electrons, the lower the resistance. The FEWER free electrons the higher the resistance.

Here' the kicker. Ready? More and Less are probably the two most important words in beginning science.

The More the resistance, the Less the current flow. That's a really important consideration in battery drain in a watch (or modern day calculator). The More the Battery Drain, the Less time it will last.

Always be careful when more and Less are around.

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Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec

SOVA2 [1]

Answer:

$F_{net} = 6.879\times 10^{- 7}\ N$

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

$F = \frac{GMm}{\frac({d}{2}^{2})}$

$F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N$

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

$F_{net} = F + F'$

$F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N$

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3 years ago

A battery establishes a voltageV on a parallel-plate capacitor. After the battery is disconnected, the capacitor doesn't loss ch

Digiron [165]

Answer:

Explanation:

I me and my friends did a project on this

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2 years ago

What is the energy (in ev) of an x-ray photon that has a wavelength of 3.1 nm ?

shepuryov [24]

Veiyxvq;ouzb1qxzbqioxzv;ux

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3 years ago

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and th

Evgen [1.6K]

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

We consider merry go-round as a ring:

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

Moment of inertia of the person considering as a point mass:

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

Now according to the conservation of angular momentum:

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$

4 0

3 years ago

Read 2 more answers

The center of the galaxy is filled with low-density hydrogen gas that scatters light rays. An astronomer wants to take a picture

Mandarinka [93]

Answer:

Infrared light

Explanation:

Infrared light is the spectrum of electromagnetic wave given off by a body possessing thermal energy. Infrared light is preferred over visible light in this region of space because visible light is easily scattered in the presence of fine particles. Infrared ray makes it easy for us to observe Cold, dark molecular clouds of gas and dust in our galaxy that glows when irradiated by the stars . Infrared can also be used to detect young forming stars, even before they begin to emit visible light. Stars emit a smaller portion of their energy in the infrared spectrum, so nearby cool objects such as planets can be more readily detected with infrared light which won't be possible with an ultraviolet or visible light.

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2 years ago