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horsena [70]
3 years ago
7

People are constantly being barraged with low-level radiation. Which radiation source below contributes the most low-level radia

tion exposure to the average person?a) cosmic raysb) medical X-raysc) radond) rocks, soils, and food
Physics
2 answers:
Gennadij [26K]3 years ago
7 0

Answer:

C) Radon

Explanation:

Humans are terrified of radiation attack but we are unaware that we face radiations from a lot of natural sources which includes food and water. But their level is low and thus do not have any major impact on our body.

As per US Nuclear Regulatory Commission,on an average, an American faces a radiation of 620 millirem each year. Out of this 50% is because of Natural background radiation. Majority of this radiation is due to Radon in the air, with small contribution (around 30 millirem) from Cosmic rays and even smaller contribution from Earth itself.

Sati [7]3 years ago
3 0

I think the answer is C) Radond

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Two astronauts are playing catch in a zero gravitational field. Astronaut 1 of mass m1 is initially moving to the right with spe
Ede4ka [16]

The final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.

The given parameters;

  • Mass of the first astronaut, = m₁
  • Mass of the second astronaut, = m₂
  • Initial velocity of the first astronaut, = v₁
  • Initial velocity of the second astronaut, = v₂ > v₁
  • Mass of the ball, = m
  • Speed of the ball, = u
  • Final velocity of the first astronaut, = v_f_1
  • Final velocity of the second astronaut, = v_f_2

The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.

m_1v_1 + m_2v_2 = m_2v_2_f + m_1v_1_f

if v₂ > v₁, then v_1_f > v_2_f, to conserve the linear momentum.

Thus, the final velocity (v_1_f) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (v_2_f) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.

Learn more here: brainly.com/question/24424291

5 0
3 years ago
A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and
icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

  • g = acceleration due to gravity = 9.8 m/s²,
  • r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
  • ρ = density of sphere = 19 g/cm³,
  • σ = density of water = 1.0 g/cm³ and
  • η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

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4 0
1 year ago
Could a person at the south pole see the north star, explain?? ​
9966 [12]

Answer:

If conditions are just right, you can see Polaris from just south of the equator.  Although Polaris is also known as the North Star, it doesn't lie precisely above Earth's North Pole. If it did, Polaris would have a declination of exactly 90 degree.

Explanation:

4 0
2 years ago
Read 2 more answers
Visible light passes through a diffraction grating that has 900 slits per centimeter, and the interference pattern is observed o
kobusy [5.1K]

Answer:

\Delta \lambda=14.3\ nm

Explanation:

It is given that,

The number of lines per unit length, N = 900 slits per cm

Distance between the formed pattern and the grating, l = 2.3 m

n the first-order spectrum, maxima for two different wavelengths are separated on the screen by 2.98 mm, \Delta Y=2.98\ mm = 0.00298\ m

Let d is the slit width of the grating,

d=\dfrac{1}{N}

d=\dfrac{1}{900\ cm}

d=1.11\times 10^{-5}\ m

For the first wavelength, the position of maxima is given by :

y_1=\dfrac{L\lambda_1}{d}

For the other wavelength, the position of maxima is given by :

y_2=\dfrac{L\lambda_2}{d}

So,

\Delta \lambda=\dfrac{\Delta y d}{l}

\Delta \lambda=\dfrac{0.00298\times 1.11\times 10^{-5}}{2.3}

\Delta \lambda=1.43\times 10^{-8}\ m

or

\Delta \lambda=14.3\ nm

So, the difference between these wavelengths is 14.3 nm. Hence, this is the required solution.

3 0
3 years ago
A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 13° above the horizontal. (a) if
Blababa [14]
<span>Answer: Therefore, x component: Tcos(24°) - f = 0 y component: N + Tsin(24°) - mg = 0 The two equations I get from this are: f = Tcos(24°) N = mg - Tsin(24°) In order for the crate to move, the friction force has to be greater than the normal force multiplied by the static coefficient, so... Tcos(24°) = 0.47 * (mg - Tsin(24°)) From all that I can get the equation I need for the tension, which, after some algebraic manipulation, yields: T = (mg * static coefficient) / (cos(24°) + sin(24°) * static coefficient) Then plugging in the values... T = 283.52. Reference https://www.physicsforums.com/threads/difficulty-with-force-problems-involving-friction.111768/</span>
7 0
3 years ago
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