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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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The repulsive force between two protons has a magnitude of 2.00 N. What is the distance between them?

Aloiza [94]

Answer:

The answer to your question is letter A. r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

$F = K\frac{q1q2}{r^{2}}$

Solve for r

$r = \sqrt{\frac{kq1q2}{F}}$

Substitution

$r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}$

Simplification

r = $\sqrt{\frac{2.3 x 10^{-28}}{2}}$

r = $\sqrt{1.15 x 10^{-24}}$

Result

r = 1.07 x 10⁻¹⁴ m

6 0

3 years ago

Read 2 more answers

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

3 0

3 years ago

State a Newtowns second law

astraxan [27]

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

5 0

3 years ago

Number of conducting plates of a multiplate capacitor is 5. The no. Of capacitors is

Ivahew [28]

Answer:

4 capacitors

Explanation:

Given

$n = 5$ --- conducting plates