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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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How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five
Keith_Richards [23]

Answer:

<em>The kinetic energy of a spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

<em></em>

Explanation:

Let us first consider the initial characteristics of the angular motion of the disk

moment of inertia = I

angular speed = ω

For the second case, we consider the characteristics to now be

moment of inertia = 5I  (five times larger)

angular speed = ω/5  (five times smaller)

Recall that the kinetic energy of a spinning body is given as

KE = \frac{1}{2}Iw^{2}

therefore,

for the first case, the K.E. is given as

KE = \frac{1}{2}Iw^{2}

and for the second case, the K.E. is given as

KE = \frac{1}{2}(5I)(\frac{w}{5} )^{2}   = \frac{5}{50}Iw^{2}

KE = \frac{1}{10}Iw^{2}

<em>this is one-tenth the kinetic energy before its spinning characteristics were changed.</em>

<em>This implies that the kinetic energy of the spinning disk will be reduced to a tenth of its initial kinetic energy if its moment of inertia is made five times larger, but its angular speed is made five times smaller.</em>

6 0
3 years ago
A train has a constant velocity of 2 m/s. east what is the magnitude of the horizontal acceleration of the trian?
kupik [55]

Answer:

0 m/s²

Explanation:

The velocity is constant, so there is no acceleration.

4 0
3 years ago
An electroplating solution is made up of nickel(II) sulfate. How much time would it take to deposit 0.500 g of metallic nickel o
ELEN [110]

Answer:

1.52 hour

Explanation:

M = 0.5 g, I = 3 A

Electrochemical equivalent of nickel

Z = 3.04 × 10^(-4) g/C

By use of Faraday's laws of electrolysis

M = Z I t

t = M / Z I

t = 0.5 / (3.04 × 10^-4 × 3)

t = 5482.45 second = 1.52 hour

6 0
3 years ago
An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 43.0 km . To the
GalinKa [24]

Answer: v₂ = 5962 km

the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

Explanation:

Given that;

Lunar surface is in an altitude h = 43.0 km =  43 × 10³ m

we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²

speed of the space craft when it crashes into the lunar surface , v

decreasing speed of the space craft = 23 m/s

Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r

but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²

HERE r = Rm + h

we substitute

r = 1.74 × 10⁶ m + 43 × 10³ m

= 1.783 × 10⁶ m

On equating these, we have

G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)

v²/r = GM/r²

v = √ ( GM/r)

v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )

v = √ (2750787.9978)

v = 1658.55 m/s

Now since speed is decreasing by 23 m/s

the speed of the space craft into the lunar face is,

v₁ = 1658.55 m/s - 23 m/s

v₁ = 1635.55 m/s

Now applying conversation of energy, we say

1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)

v₂ =  √ [ v₁² + GMe (1/Rm - 1/r)]

v₂ =   √ [ 1635.55²  + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]

v₂ =  √ (2675023.8025 + 67979.24)

v₂ = √(2743003.046)

v₂ = 1656.2 m/s

now convert

v₂ = 1656.2 × 1km/1000m × 3600s/1hrs

v₂ = 5962 km

Therefore the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

8 0
3 years ago
Estimate the kinetic energy (in GJ) of a 98,000 metric ton aircraft carrier moving at a speed of at 32 knots. You will need to l
kiruha [24]

Answer:

K = 13.276 GJ

Explanation:

Given:

Mass of the aircraft (m) = 98,000 metric ton

Speed of the aircraft (v) = 32 knots

Kinetic energy of the aircraft (K) = ?

We know that,

1 metric ton = 1000 kg

∴ 98,000 metric ton = 98,000 × 1000 = 9.8 × 10⁷ kg

Also, 1 knot = 1 nautical mile per hour

So, 32 knots = 32 nautical miles per hour

Converting nautical miles per hour to meter per second using the following conversions, we get:

1 nautical mile = 1852 m

1 hour = 3600 s

Therefore, 1 nautical mile per hour = \frac{1852}{3600}\ m/s

So, 32 nautical miles per hour = 32\times \frac{1852}{3600}=16.46\ m/s

Therefore, the mass and speed of the aircraft are:

m=9.8\times 10^{7}\ kg\\v=16.46\ m/s

Now, kinetic energy is given as:

K=\frac{1}{2}mv^2

Plug in the above values and solve for 'K'. This gives,

K=\frac{1}{2}\times 9.8\times 10^7\times (16.46)^2\\\\K=4.9\times 270.9316\times 10^7\ J\\\\K=13.276\times 10^2\times 10^7\ J\\\\K=13.276\times 10^9\ J

Now, we know that, 1 GJ = 10⁹ J

Therefore, the value of 'K' in terms of GJ is given as:

K = 13.276 GJ

8 0
3 years ago
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