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natta225 [31]
3 years ago
10

A ray of light has a wavelength of

Physics
1 answer:
MArishka [77]3 years ago
6 0

Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, \lambda=284\ nm

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm

So, the wavelength in vacuum is equal to 428.8 nm.

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A person walks 24 meters at 3 m/s. How long did it take them
asambeis [7]

Hi my dear friend,

2m*30 = 60m

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4 0
3 years ago
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Milk with a density of 1020 kg/m3 is transported on a level road in a 9-m-long, 3-m-diameter cylindrical tanker. The tanker is c
jeka57 [31]

Solution :

Given data is :

Density of the milk in the tank, $\rho = 1020 \ kg/m^3$

Length of the tank, x = 9 m

Height of the tank, z = 3 m

Acceleration of the tank, $a_x = 2.5 \ m/s^2$

Therefore, the pressure difference between the two points is given by :

$P_2-P_1 = -\rho a_x x - \rho(g+a)z$

Since the tank is completely filled with milk, the vertical acceleration is $a_z = 0$

$P_2-P_1 = -\rho a_x x- \rho g z$

Therefore substituting, we get

$P_2-P_1=-(1020 \times 2.5 \times 7) - (1020 \times 9.81 \times 3)$

           $=-17850 - 30018.6$

           $=-47868.6 \ Pa$

           $=-47.868 \ kPa$

Therefore the maximum pressure difference in the tank is Δp = 47.87 kPa and is located at the bottom of the tank.

         

4 0
2 years ago
What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency
Anvisha [2.4K]

Answer:

1.04\times 10^{-7} T

Explanation:

IP  = Power of the bulb = 100 W

r  = distance from the bulb = 2.5 m

I = Intensity of light at the location

Intensity of the light at the location is given as

I = \frac{P}{4\pi r^{2}}

I = \frac{100}{4(3.14) (2.5)^{2}}

I = 1.28 W/m²

B_{o} = maximum magnetic field

Intensity is given as

I = \frac{B_{o}^{2}c}{2\mu _{o}}

1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}

B_{o} = 1.04\times 10^{-7} T

7 0
3 years ago
In a cyclic process, a gas performs 123 J of work on its surroundings per cycle. What amount of heat, if any, transfers into or
Margaret [11]

Answer:

123 J transfer into the gas

Explanation:

Here we know that 123 J work is done by the gas on its surrounding

So here gas is doing work against external forces

Now for cyclic process we know that

\Delta U = 0

so from 1st law of thermodynamics we have

dQ = W + \Delta U

dQ = W

so work done is same as the heat supplied to the system

So correct answer is

123 J transfer into the gas

8 0
3 years ago
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