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GenaCL600 [577]
3 years ago
15

It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans), using x-rays, just

to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 mSv, applied to the whole body. By contrast, a chest x-ray typically administers 0.20 mSv to only 6.0 kg of tissue. How many chest x rays would deliver the same total amount of energy to the body of a 85 kg person as one whole-body scan?
Physics
1 answer:
aksik [14]3 years ago
7 0

Answer:

Explanation:

Given

One CT scan gives a dose of 12 mSv

and a chest X ray typically administer 0.2 mSv to 6 kg tissue

mass of Person m=85 kg

therefore \frac{85}{6}=14.167

N no of x-Ray equals to 12 mSv dose of CT scan

n=\frac{12}{0.2}=60

Therefore total no of x ray required is =60\times 14.167=850.02\approx 850 X rays

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2. A powerful experimental sewing machine is powered by a mass-spring system. This
Alexus [3.1K]

We have that the Number of stitches per sec and he mass of  oscillation motion is mathematically given as

a) Nt=25stitches per sec

b) m=2.033e-5kg

<h3>Number of stitches per sec and he mass of  oscillation motion</h3>

Question Parameters:

This <u>sewing </u>machine is capable of stitching 1,500 stiches in one minute.

If the <em>sewing </em>machine has a spring constant of 0.5 N/m,

Generally the equation for the Number of stitches per sec  is mathematically given as

Nt=N/t

Therefore

Nt=1500/60

Nt=25stitches per sec

b)

Generally the equation for the Time t  is mathematically given as

T=2\pi\sqrt{\frac{m}{k}}

Therefore

0.04=2\pi\sqrt{\frac{m}{0.5}}\\\\m=\frac{0.5*0.04^2}{4\pi^2}

m=2.033e-5kg

For more information on Mass visit

brainly.com/question/15959704

7 0
2 years ago
What could you do to change the volume of a gas?
Lorico [155]
The ONLY way to change the volume of a sample of gas is to transfer it to a container with different volume.
Simply changing its temperature or pressure in the same jar won't do it. Any amount of gas always fills whatever container you keep it in.
8 0
3 years ago
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
Read 2 more answers
What grade of sprain is a completely torn ligament?
Alex Ar [27]
Grade 1: Stretching or slight tearing of the ligament with mild tenderness, swelling and stiffness. The ankle feels stable and it is usually possible to walk with minimal pain.

Grade 2: A more severe sprain, but incomplete tear with moderate pain, swelling and bruising. Although it feels somewhat stable, the damaged areas are tender to the touch and walking is painful.

Grade 3: This is a complete tear of the affected ligament(s) with severe swelling and bruising. The ankle is unstable and walking is likely not possible because the ankle gives out and there is intense pain.

source - https://www.rushcopley.com/health/physician-articles/varying-degrees-of-ankle-sprains/
6 0
3 years ago
Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.
azamat

Answer:

\theta=5.71^{o}

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

\sum F_{y}=0

-W_{y}+N=0

N=W_{y}

so:

N=mgcos(\theta)

now we can go ahead and do a sum of forces in the x-direction:

\sum F_{x}=0

the sum of forces in x is 0 because it's moving at a constant speed.

-f+W_{x}=0

-\mu_{k}N+mg sen(\theta)=0

-\mu_{k}mg cos(\theta)+mg sen(\theta)=0

so now we solve for theta. We can start by factoring mg so we get:

mg(-\mu_{k} cos(\theta)+sen(\theta))=0

we can divide both sides into mg so we get:

-\mu_{k} cos(\theta)+sen(\theta)=0

this tells us that the problem is independent of the mass of the object.

\mu_{k} cos(\theta)=sen(\theta)

we now divide both sides of the equation into cos(\theta) so we get:

\mu_{k}=\frac{sen(\theta)}{cos(\theta)}

\mu_{k}=tan(\theta)

so we now take the inverse function of tan to get:

\theta=tan^{-1}(\mu_{k})

so now we can find our angle:

\theta=tan^{-1}(0.10)

so

\theta=5.71^{o}

8 0
3 years ago
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