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2 years ago

What if a wave reaches a type of

2 answers:

7 0

If it hit the matter that it cannot pass through.....according to the velocity of wave..it might get reflected back....we can find sound waves reflecting back in closed rooms...because of the presence of the matter through which it cannot pass....and...it's because of this tht...echo is observed

umka21 [38]2 years ago

5 0

It is bounced back. Like echoing, it's basically the sound wave hitting the matter it can't go through and bounce back.

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In a two-slit experiment using coherent light, the distance between the slits and the screen is 1.10 m, and the distance between

Paul [167]

Answer:

D) 763 nm

Explanation:

Calculation for the wavelength of light

Using this formula

Wavelength of light=Delta Y*Distance / Length

Where,

Delta Y represent the 2nd order bright fringe

Length represent the distance between both the slits and the screen

Distance represent the Distance between the slits

Let note that cm to m = (4.2) x 10^-2 and mm to m= ( 0.0400x 10^-3)

Now Let plug in the formula

Wavelength of light=[(4.2 x 10^-2m)(0.0400 x 10^-3m) / 2(1.1m)]*10^-7 meters

Wavelength of light=[(0.042m) (0.0004m)/2.2m]*10^-7 meters

Wavelength of light =(0.0000168m/2.2m)*10^-7 meters

Wavelength of light =7.63 *10^-7 meters

Wavelength of light =763 nm

Therefore the Wavelength of light will be 763 nm

3 0

2 years ago

Suppose a clay model of a koala bear has a mass of 0.200 kg and slides on ice at a speed of 0.750 m/s. It runs into another clay

kaheart [24]

Answer:

0.278 m/s

Explanation:

We can answer the problem by using the law of conservation of momentum. In fact, the total momentum before the collision must be equal to the total momentum after the collision.

So we can write:

$mu=(m+M)v$

where

m = 0.200 kg is the mass of the koala bear

u = 0.750 m/s is the initial velocity of the koala bear

M = 0.350 kg is the mass of the other clay model

v is their final combined velocity

Solving the equation for v, we get

$v=\frac{mu}{m+M}=\frac{(0.200)(0.750)}{0.200+0.350}=0.278 m/s$

8 0

3 years ago

When exposed to a radioactive source which emits 1.2-MeV gamma-rays, a particular material is found to have a half-value thickne

iren2701 [21]

Answer:E

Explanation:

It is given that Energy of gamma ray is E=1.2 Mev

Shielding effect can be measured by measuring the fraction of gamma rays blocked by shield. If certain thickness will able to block half the radiation then to block 75% radiation we need to add same amount of thickness in order to block the remaining radiation.

i.e. $\frac{E}{2}$ fraction is blocked by 10 cm thickness

then remaining radiation is $\frac{E}{2}$

another 10 cm thickness will block the remaining half radiation i.e. $\frac{1}{2}\times \frac{E}{2}=\frac{E}{4}$

so total 75 % radiation will be blocked

4 0

3 years ago

What is the relationship between thermal energy, temperature, pressure?

Ira Lisetskai [31]

Answer: When volume is constant, pressure is directly proportional to temperature. When temperature is constant, pressure is inversely proportional to volume. When pressure is constant, volume is directly proportional to temperature.

Explanation:

6 0

3 years ago

A deuteron, 21H, is the nucleus of a hydrogen isotope and consists of one proton and one neutron. The plasma of deuterons in a n

Naddik [55]

Answer:

1917723.40119 m/s

Explanation:

m = Mass of deuteron = $(1.637+1.675)\times 10^{-27}=3.312\times 10^{-27}$

k = Boltzmann constant = $1.381\times 10^{-23}\ J/K$

T = Temperature = $2.94\times 10^8\ K$

RMS velocity is given by

$V_r=\sqrt{\dfrac{3kT}{m}}\\\Rightarrow V_r=\sqrt{\dfrac{3\times 1.381\times 10^{-23}\times 2.94\times 10^8}{3.312\times 10^{-27}}}\\\Rightarrow V_r=1917723.40119\ m/s$

The RMS velocity of the deutrons is 1917723.40119 m/s

8 0

3 years ago