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3 years ago

Evidence of wave theory light

1 answer:

7 0

All of these things were seen in formal experiments by the 19th century. But some of them are easy to see in your own home. It's obvious that light can reflect - you just have to look in a mirror. Light bounces off the mirror and goes into your eye so you can see yourself. It's also obvious that light can refract: All you have to do is put a spoon in a large glass of water and watch how the spoon appears to bend.

That happens because the light is bending as it moves between air and water. Both of these things can be seen even more clearly in a laboratory using beams of light or lasers.

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In a science museum, a 140 kg brass pendulum bob swings at the end of a 16.8 m -long wire.

Mice21 [21]

The period of the pendulum is 8.2 s

Explanation:

The period of a simple pendulum is given by the equation:

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration of gravity

T is the period

We notice that the period of a pendulum does not depend at all on its mass, but only on its length.

For the pendulum in this problem, we have

L = 16.8 m

and

$g=9.8 m/s^2$ (acceleration of gravity)

Therefore the period of this pendulum is

$T=2\pi \sqrt{\frac{16.8}{9.8}}=8.2 s$

#LearnWithBrainly

3 0

3 years ago

If you touch a hot light bulb and get burned, that is an example of _____.

Anton [14]

The answer is C conduction

7 0

2 years ago

Read 2 more answers

A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, $\alpha=3.3 rad/s^2$

Diameter of the wheel, d=21 cm

Radius of wheel, $r=\frac{d}{2}=\frac{21}{2}$ cm

Radius of wheel, $r=\frac{21\times 10^{-2}}{2} m$

1m=100 cm

Magnitude of total linear acceleration, a= $1.7 m/s^2$

We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration, $a_t=\alpha r$

$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

$a_t=34.65\times 10^{-2}m/s^2$

Radial acceleration, $a_r=\frac{v^2}{r}$

We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

6 0

3 years ago

A 25.0-kg test rocket is fired vertically. When the engine stops firing, the rocket’s kinetic energy is 2017 J. After the fuel i

Dahasolnce [82]

Answer:

1. 2.12105 J. The final kinetic energy is. KE f mv2. (875.0 kg)(44.0 m/s)2 2.What is the velocity of the two hockey players after the collision? ... A 10.0-kg test rocket is fired vertically from.

Explanation:

Sana maka tulong

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3 years ago

Starting fom rest, a car accelerates at a constant rate, reaching 88 km/h in 12 s. (a) What is its acceleration? (b) How far doe

JulsSmile [24]

Answer:

(A) $a=2.0.37m/sec^2$