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3 years ago

Which part of the electromagnetic spectrum has a higher frequency than ultraviolet light

Physics

1 answer:

bezimeni [28]3 years ago

8 0

X-rays and gamma rays are kept up there.

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In a physics lab, light with a wavelength of 570 travels in air from a laser to a photocell in a time of 16.5 . When a slab of g

aliina [53]

Answer:

Wavelength is calculated as 213.9 nm

Solution:

As per the question:

Wavelength of light = 570 nm

Time, t = 16.5 ns

Thickness of glass slab, d = 0.865 ns

Time taken to travel from laser to the photocell, t' = 21.3

Speed of light in vacuum, c = $3\times 10^{8}\ m/s$

Now,

To calculate the wavelength of light inside the glass:

After the insertion of the glass slab into the beam, the extra time taken by light to cover a thickness t = 0.865 m is:

t' - t = 21.3 - 16.5 = 4.8 ns

Thus

$\frac{d}{\frac{c}{n}} - \frac{d}{v} = 4.8\times 10^{- 9}$

$\frac{0.8656}{\frac{c}{n}} - \frac{0.865}{v} = 4.8\times 10^{- 9}$

where

n = refractive index of the medium

v = speed of light in medium

$\frac{0.8656}{\frac{c}{n}} - \frac{0.865}{v} = 4.8\times 10^{- 9}$

$n = \frac{4.8\times 10^{- 9}\times 3.00\times 10^{8}}{0.865} + 1$

n = 2.66

Now,

The wavelength in the glass:

$\lambda' = \frac{\lambda }{n}$

$\lambda' = \frac{570}{2.66} = 213.9\ nm$

5 0

3 years ago

Elaborate on the reason(s) that matter is said to move even as in a solid state.

Mariulka [41]

<u>Answer:</u>

The matter does not move in solid state but vibrates.

<u>Explanation:</u>

The atoms inside the matter cannot move or shift their positions without any external force but makes some small vibration movements. Generally in solids, the particles are bound by the attractive forces acting in between the atoms inside the matter.

The small vibrations that are happening inside the matter are because of the external factors like temperature. The increase in temperature raises the kinetic energy of the atoms inside and makes them move faster and this results in the vibration of the matter.

6 0

3 years ago

Read 2 more answers

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

3 years ago

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

5 0

3 years ago

Suppose that the Sun started shrinking in size, without losing any mass. What would be the effect of the Sun's change on the orb

adell [148]

Answer:

F = G M m / R^2 gravitational force on planet of mass m.

None of these quantities change in the given hypothesis so

there will be no change in the orbit of mass m

3 0

2 years ago