What is the smallest height needed to make a wave?

In two separate double slit experiments, an interference pattern is observed on a screen. In the first experiment, violet light

sashaice [31]

Answer:

$\lambda_3 = 4.72*10^{-7} m$

Explanation:

given data:

wavelength \lambda = 708nm = 708*10^{-9} m

using the following relation:

$y = \frac{mL\lambda}{d}$

according to the given information

second and third dark fringe is at same location. so

$y_2 = y_3$

$\frac{m_2L\lambda_2}{d} = \frac{m_3L\lambda_3}{d}$

$m_2\lambda_2 = m_3\lambda_3$

$2*708*10^{-9} = 3*\lambda_3$

$\lambda_3 = \frac{2*708*10^{-9}}{3}$

$\lambda_3 = 4.72*10^{-7} m$

4 0

2 years ago

A rigid vessel of 0.06 m3 volume contains an ideal gas , CV =2.5R, at 500K and 1 bar.a). if 15000J heat is transferred to the ga

andreev551 [17]

Answer:

Given that

V= 0.06 m³

Cv= 2.5 R= 5/2 R

T₁=500 K

P₁=1 bar

Heat addition = 15000 J

We know that heat addition at constant volume process ( rigid vessel ) given as

Q = n Cv ΔT

We know that

P V = n R T

n=PV/RT

n= (100 x 0.06)(500 x 8.314)

n=1.443 mol

So

Q = n Cv ΔT

15000 = 1.433 x 2.5 x 8.314 ( T₂-500)

T₂=1000.12 K

We know that at constant volume process

P₂/P₁=T₂/T₁

P₂/1 = 1000.21/500

P₂= 2 bar

Entropy change given as

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

Cp-Cv= R

Cp=7/2 R

Now by putting the values

$\Delta S=nC_P\ln \dfrac{T_2}{T_1}-nR\ln \dfrac{P_2}{P_1}$

$\Delta S=1.443\times 3.5\times 8.314\ln \dfrac{1000.21}{500}-1.443\times 8.314\ln \dfrac{2}{1}$

a)ΔS= 20.79 J/K

b)

If the process is adiabatic it means that heat transfer is zero.

So

ΔS= 20.79 J/K

We know that

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

Process is adiabatic

$\Delta S_{surr}=0$

$\Delta S_{univ}=\Delta S_{syatem}+\Delta S_{surr}$

$\Delta S_{univ}= 20.79 +0$

$\Delta S_{univ}= 20.79$

3 0

2 years ago

A child's toy consists of a m = 36 g monkey suspended from a spring of negligible mass and spring constant k. When the toy monke

kolezko [41]

Answer:

Part A - 3N/m

Part B - see attachment

Part C - 4.9 × 10-³J

Part D - E = 1/2kd² + 1/2mv² + mgh

Explanation:

This problem requires the knowledge of simple harmonic motion for cimplete solution. To find the spring constant in part A the expression relating the force applied to a spring and the resulting stretching of the spring (hooke's law) is required which is F = kx.

The free body diagram can be found in the attachment. Fp(force of pull), Ft(Force of tension) and W(weight).

The energy stored in the pring as a result of the stretching of d = 5.7cm is 1/2kd².

Part D

Three forces act on the spring-monkey system and they do work in different forms: kinetic energy 1/2mv² , elastic potential

energy due to the restoring force in the spring or the tension force 1/2kd², and the gravitational potential energy mgh of the position of the system. So the total energy of the system E = 1/2kd² + 1/2mv² + mgh.

8 0

2 years ago

There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which

Dmitrij [34]

Answer: B

Explanation:

Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.

4 0

2 years ago

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

Viktor [21]

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.