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3 years ago

When a burning stick of increase is moved fast in a circle a circle of red light is seen.

Physics

1 answer:

anzhelika [568]3 years ago

7 0

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

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Calculate the force of gravity on a 1–kilogram box located at a point 1.3 × 107 meters from the center of Earth if the force on

Sati [7]

Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.

Look at the formula for the gravitational force:

F = G m₁ m₂ / R² .

If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:

(6.1 N) / (2.5 kg) = (F) / (1 kg)

Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)

Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .

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4 years ago

Oceanic water particles mainly move in circles; is this movement greater on the ocean's surface or below the surface? Explain yo

goblinko [34]

I think that the oceanic water particles mainly move in circles greater in the oceans surface because of how big the waves can be and how wind and air impact the motion. The water particles move more on the surface because of the other factors that impact it such as people, wind, air, etc...

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3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

What will it be ? Plz help me

IgorC [24]

Answer:

For number 2, it will be 8.

3 0

3 years ago

Each bat can produce different frequencies of ultrasound. Explain why this is useful if there are many bats hunting together.

Crazy boy [7]

Answer: Although low frequency sound travels further than high-frequency sound, calls at higher frequencies give the bats more detailed information--such as size, range, position, speed and direction of a prey's flight. Thus, these sounds are used more often.

Explanation:

6 0

3 years ago

When a burning stick of increase is moved fast in a circle a circle of red light is seen.​

When a burning stick of increase is moved fast in a circle a circle of red light is seen.