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bixtya [17]
3 years ago
5

A student is concerned about background radiation that she is exposed to. She claims that nuclear power plants have caused backg

round radiation, and that if they had not been built, she would not be exposed to radioactivity. What is incorrect about her reasoning
Physics
2 answers:
andreev551 [17]3 years ago
7 0

The answer is D) No, because background radiation comes from naturally occurring substances such as isotopes.

Explanation:

I took it answer c is wrong

denis-greek [22]3 years ago
5 0
Based on the given statement or reasoning above made by a student, what makes her reasoning incorrect is the part where she mentioned that "if they had not been built, she would not be exposed to radioactivity." This is definitely false. What makes this incorrect is that background radiation is a kind of radiation that is emitted by the sun, stars, and other bodies outside the universe and therefore, no one can escape from this. Hope this helps.
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How long would it take a machine to do 5.000
Gala2k [10]

Answer:

a

Explanation:

How long would it take a machine to do 5.000

joules of work if the power rating of the machine

is 100 watts?

6 0
3 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
Nuetrik [128]

Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
3 years ago
You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?
Natali5045456 [20]
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

y_{f}=y_{o}+0+ \frac{1}{2} at^{2}

y_{f}=y_{o}+\frac{1}{2} at^{2}

y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

a = g = 9.81  \frac{m}{ s^{2}}

\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
4 0
3 years ago
Read 2 more answers
A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full
AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

A=\pi r^2

Put the value into the formula

A=\pi\times21^2

A=441\pi\ feet^2

Thickness of slice t=\Delta h\ ft

The volume is,

V=(441\pi\times\Delta h)\ ft^3

We need to calculate the force

Using formula of force

F=W\times V

Where, W = water weight

V = volume

Put the value into the formula

F=62.5\times(441\pi\times\Delta h)

F=27562.5\pi\times\Delta h\ lbs

We need to calculate the work done

Using formula of work done

W=F\times d

Put the value into the formula

W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh

W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(10\times10-\dfrac{100}{2}-0)

W=4329507.37572

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

W=\int_{0}^{h}{W}

Put the value into the formula

W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh

W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}

W=27562.5\pi(12\times10-\dfrac{100}{2}-0)

W=1929375\pi

W=6061310.32\ foot- pound

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is  is 6061310.32 foot-pound.

8 0
3 years ago
Vapor pressure is related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the _____. i
iVinArrow [24]

Answer

-Directly;  outside air pressure

Vapor pressure is directly related to the temperature of the liquid. user: in an open system, the vapor pressure is equal to the outside air pressure.

Explanation;

-As the temperature of a system increases, the average kinetic energy of the molecules increases in both the liquid and gas phases.

-A higher average kinetic energy facilitates the escape of molecules from the liquid phase into the gas phase. At the same time, the rate of return of gas phase molecules to the liquid also increases. A new equilibrium point is reached at a higher gaseous vapor pressure. The increase in vapor pressure with temperature is exponential.


6 0
3 years ago
Read 2 more answers
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