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3 years ago

Why can you see an electric bulb at night when it is switched on ?

Physics

1 answer:

Dominik [7]3 years ago

8 0

Answer:

Explanation:

Because when the electric bulb at night switched It emitted the light this light rays forms an image on our retina hence the bulb become visible at night

2 ans

When an electric bulb is switched on, the electricity is allowed to pass in its circuit and is converted into light energy. Thus, a bulb glows when switched on and the eyes perceives the light energy in the visible spectra which is given out by the bulb. Hence, the bulb becomes visible

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Approximately how long is a light-year?

Roman55 [17]

C. is the correct response.

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3 years ago

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An object of mass m is traveling around a circle of radius v. What happens to the centripetal force if the radius is cut in half

german

Answer:

Centripetal force doubles

Explanation:

Since centripetal force is given by

$F=\frac {mv^{2}}{r}$

Where F is the centripetal force, m is the mass, v is the centripetal velocity and r is circle radius.

When the radius is reduced to half then the centripetal force will be

$F=\frac {mv^{2}}{0.5r}=\frac {2mv^{2}}{r}$

Evidently, the centripetal force is doubled compared to the initial.

Note that while the question gives letter v for radius, I used v for velocity and r for radius since these are the standard letters for abbreviating formula of centripetal force.

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3 years ago

What is newton's second law of motion?

KATRIN_1 [288]

<span>the acceleration of an object is depedent upon two variables

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4 0

3 years ago

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An Olympian swims at a rate of 10.8 Km/h. How far will he travel after 0.3 hours?

zalisa [80]

Answer:

C. 3.2km

Explanation:

Seeing as the swimmer travels 10.8 km in an hour, and they are asking you to find how far they have traveled after 0.3 hours, multiply 10.8 km/h by 0.3 h. The h's will cancel and you are left with 3.24km, which can be rounded to 3.2km. Hope this helps! :)

6 0

3 years ago

Show which of the following functional forms work or do not work as solutions to this differential equation (known as 'the wave

Akimi4 [234]

Answer:

E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

Explanation:

The wave equation is

d²y / dx² = 1 /v² d²y / dt²

Where v is the speed of the wave

Let's review the solutions

In this case y = E

a) E = A sin Bt

This cannot be a solution because the part in x is missing

dE / dx = 0

b) E = A cos (Bt + c)

There is no solution missing the Part in x

dE / dx = 0

c) E = A x² t²

The parts are fine, but this solution is not an oscillating wave, so it is not an acceptable solution of the wave equation

d) E = A cos (Bx + Ct) and E = A cos (Bx + Ct + D)

Both are very similar

Let's make the derivatives

dE / dx = -A B sin (Bx + Ct + D)

d²E / dx² = -A B² cos (Bx + Ct + D)

dE / dt = - A C sin (Bx + Ct + D)

d²E / dt² = -A C² cos (Bx + Ct + D)

We substitute in the wave equation

-A B² cos (Bx + Ct + D) = 1 / v² (-A C² cos (Bx + Ct + D))

B² = 1 / v² (C²)

v² = (C / B)²

v = C / B

We see that the two equations can be a solution to the wave equation, the last one is the slightly more general solution

8 0

3 years ago