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3 years ago

In a circuit, what is responsible for lighting a bulb?

Physics

2 answers:

Yakvenalex [24]3 years ago

6 0

Answer:

3) flow of electrons, if im correct

icang [17]3 years ago

6 0

Answer:

Explanation:

The flow of electrons. A current means the flow of electrons and without a current there would be no electricity therefore the bulb would not light up.

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Please help me on these questions in the picture.

inessss [21]

2. Because it has gravity and gravity pulls it to the earth.
3. Because of the chemicals it has in it.
4. a. by running with it
    b. by pulling it.
    c. by turning it.
    d. by truning it.

3 0

3 years ago

Read 2 more answers

An object that has momentum must have

Mama L [17]

Answer:

Mechanical energy

Explanation:

If an object does not have any momentum, then it doesn't have mechicanl energy.

3 0

3 years ago

What kind of motion is the motion of the “flying bully”? How much and what is

lesantik [10]

<span>The flying bully is a move used in the Superhero Movie "Hancock", it is not a real motion in our universe. However, the direction would be towards the target object and the acceleration would be maximal.</span>

7 0

4 years ago

we measure a voltage difference of 5.0 V between two points on the conducting paper between two parallel conducting electrodes.

Alex17521 [72]

Answer:

E=1824.81 V/m

Explanation:

Given that

Voltage difference = 5 V

Distance ,D= 3 mm

θ = 24°

As we know that electric filed given as

$E=\dfrac{V}{d}$

Given that D is 24° with respect to the perpendicular to the electrodes.So we have to take cos component of D.

d= D cosθ

d= 3 cos24°

d = 2.74 mm

$E=\dfrac{V}{d}$

$E=\dfrac{5}{2.74\times 10^{-3}}$

E=1824.81 V/m

8 0

3 years ago

A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$

6 0

3 years ago