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solong [7]
3 years ago
11

Pls help, it’s due tonight :(

Physics
1 answer:
Vesnalui [34]3 years ago
6 0

Answer:

it is a decreased risk because snow blocks many ultra v rays

Explanation:

You might be interested in
the earth is broken into smaller subsystems including the atmosphere the biosphere and the hydrosphere true or false
Romashka [77]

Answer:

true

Explanation:

these are not the only parts of the atmosphere, i dont know the full list but i know these arent the only parts

7 0
3 years ago
You set a tuning fork into vibration at a frequency of 723 Hz and then drop it off the roof of the Physics building where the ac
zaharov [31]

Answer:

Explanation:

Given

Original Frequency f=723\ Hz

apparent Frequency f'=697\ Hz

There is change in frequency whenever source move relative to the observer.

From Doppler effect we can write as

f'=f\cdot \frac{v-v_o}{v+v_s}

where  

f'=apparent frequency  

v=velocity of sound in the given media

v_s=velocity of source

v_0=velocity of observer  

here v_0=0

697=723\cdot (\frac{343-0}{343+v_s})

v_s=(\frac{f}{f'}-1)v

v_s=(\frac{723}{697}-1)\cdot 343

v_s=12.79\approx 12.8\ m/s

i.e.fork acquired a velocity of 12.8 m/s

distance traveled by fork is given by

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

v_s^2-0=2\times 9.8\times s

s=\frac{12.8^2}{2\times 9.8}

s=8.35\ m

                                       

5 0
3 years ago
In a mercury barometer at atmospheric pressure, the height of the column of mercury in a glass tube is 760 mm. If another mercur
Archy [21]

Answer:

c. equal to 760 mm

Explanation:

We are told that another mercury barometer is used that has a tube of larger diameter. This means a larger area and the weight of the liquid in the tube will have increased since volume = area × height. Also, due to the larger area, the net upward force on the mercury will have also increased by the same amount because force = area × pressure. Therefore, as long as the pressure remains the same, the height of the mercury will also remain the same.

Thus the height of the mercury = 760 mm

7 0
3 years ago
A ray of light is moving from a material having a high index of refraction into a material with a lower index of refraction.(a)
Sveta_85 [38]

(a) Away from the normal

The direction of bending of the ray of light can be found by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1, n2 are the index of refraction of the first and second medium, respectively

\theta_1, \theta_2 are the angle of the incident and refracted ray with respect to the normal to the surface, respectively

In this problem, the ray of light moves from a material with high index of refraction to a material with lower index, so:

n_1 > n_2

Re-arranging Snell's law we find

sin \theta_2 = \frac{n_1}{n_2} sin \theta

since

\frac{n_1}{n_2}>1

we also have

sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength of the light in a vacuum

n is the refractive index

We can rewrite the equation as

\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2

And isolating \lambda_2 from the second equation

\lambda_2 = \frac{n_1}{n_2} \lambda_1

where

\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1

So, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency is the same

While wavelength and speed of a light wave depend on the medium in which the wave is travelling through, the frequency does not depend on that, so it is exactly the same in the two mediums.

5 0
4 years ago
Although the evidence is weak, there has been concern in recent years over possible health effects from the magnetic fields gene
Natasha2012 [34]

Answer:

A. B = 6.36 * 10^{-10} T

B. P ≈ 0

Explanation:

In order to calculate the magnetic field strength we have to use the magnetic field strength of a straight wire.

B = \frac{mi* I}{2\pi *d} (eq. I)

B = magnetic field strength at distance d

I = current (A)

mi = represented by the greek letter μ, represents the permeability of the free space, which is: 4 × π 10^(-7) T m/A

d = distance from the wire

By replacing the values in eq I, we have the following:

B = \frac{4\pi  10^{-7} T  m  A^{-1}  200 A}{2\pi *20 m}\\\\B = 6.36 * 10^{-10}  T\\ (eq II)

The earth magnetic field in the surface variates from 25 to 65 microteslas. Thus:

P = Percentage from the wires/percentage of the earth

P = \frac{6.36 * 10^{-10}T}{65* 10^{-3} T}\\ ∵ B ∴

P ≈ 0

5 0
4 years ago
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