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mina [271]
4 years ago
14

Why should photons emitted by a hotter gas have, on average, shorter wavelengths than photons emitted by a cooler gas?

Physics
1 answer:
harkovskaia [24]4 years ago
7 0

According to wein's law,

wavelength has inverse relation with temperature

so that's why hotter gas has shorter wavelength than cooler gas.

hope this answer is correct ....

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Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

7 0
4 years ago
An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons
vlada-n [284]

Answer:

v=2.02\frac{m}{s}

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}

8 0
3 years ago
a wave in a rope is traveling at 6 m/s and at a frequency of 2 Hz. what is the wavelength of the wave producced
Tems11 [23]

Answer:

3

Explanation:

4 0
3 years ago
Which one is itttttttttttt
Vlada [557]

Answer:

Red

Explanation:

6 0
3 years ago
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor sto
Fed [463]

Answer:

a)   a = 34.375 m / s²,  b)    v_f = 550 m / s

Explanation:

This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.

a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed

             v_f = \frac{x-x_1}{t}

we substitute the values

             v_f = \frac{ 6600 -x_1}{4}  

The initial part of the movement is carried out with acceleration

             v_f = v₀ + a t

             x₁ = x₀ + v₀ t + ½ a t²

the rocket starts from rest v₀ = 0 with an initial height x₀ = 0

             x₁ = ½ a t²

             v_f = a t

we substitute the values

              x₁ = 1/2  a 16²

              x₁ = 128 a

              v_f = 16 a

let's write our system of equations  

               v_f = \frac{6600 - x_1}{4}

               x₁ = 128 a

               v_f = 16 a

we substitute in the first equation  

               16 a = \frac{6600 -128 a}{4}

               16 4 a = 6600 - 128 a

                a (64 + 128) = 6600

                a = 6600/192

                 a = 34.375 m / s²

b) let's find the time to reach this height

                x = ½ to t²

                t² = 2y / a

                t² = 2 5100 / 34.375

                t² = 296.72

                t = 17.2 s

We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part

               v_f = 16 a

               v_f = 16 34.375

               v_f = 550 m / s

8 0
3 years ago
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