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vampirchik [111]
3 years ago
8

The “Compton effect” showed

Physics
2 answers:
Blizzard [7]3 years ago
8 0
The answer is the last choice, the particle nature of light. The compton effect showed that a charged electron scatters the photon results in decreased energy but increase in wavelength.
Yanka [14]3 years ago
3 0

Answer:

the particle nature of light

Explanation:

As we know that by the formula of Compton Effect we have

\lambda' - \lambda = \frac{h}{mc}(1 - cos\theta)

here in this formula we will say that when light of some given wavelength \lambda incident on a particle then after deviating from the particle the light will change its direction by some angle \theta.

And the wavelength of the particle will also changed to some different value as \lambda'

so here as per particle theory we will use the theory of momentum conservation as per which light of given momentum when hit a particle then it will transfer it momentum to the particle and deviated by some lesser momentum.

Due to decrease in momentum the change in wavelength of light is given by above formula

so this verify the particle nature of light

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A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250
Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30

coefficient of static friction \mu =0.35

coefficient of kinetic friction \mu _k=0.25

distance traveled d=77.3 cm

spring constant k=35 N/m

work done by gravity+work done by friction=Energy stored in Spring

mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}

mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}

0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}

x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}

x=0.234 m

x=23.4 cm

6 0
3 years ago
Modern nuclear bomb tests have created an extra high level of14C in our atmosphere. When future archaeologistsdate samples from
Ipatiy [6.2K]

Answer:

Too old(Ex. if real time is 1000 then they estimate >1000)

Explanation:

This is because with time our planet may have a definite function which describes temperature.(Because of all the factors and global warming except nuclear bomb testing)

Now nuclear test on planet have significant effect on temperature rise.

Also 14°C rise in temperature is good one because of this.

If future archaeologists only consider that  uniform function as above mentioned then they estimate more time then the real one.

Thus too old is right answer.

8 0
3 years ago
Density is given in ____.<br> a. Pa/cm3<br> c. g/s2<br> b. N/m2<br> d. g/cm3
Ainat [17]
Density is defined as  [mass] / [volume] .

The only choice listed with those physical dimensions is 'd' .
3 0
3 years ago
Why does of beam of light ben when it passes into a new medium?
Pachacha [2.7K]

Answer:

The bending occurs because light travels more slowly in a denser medium.

3 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
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