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aleksley [76]
3 years ago
5

Radioisotopes often emit alpha particles, beta particles, or gamma rays. The distance they travel through matter increases in or

der from alpha to gamma. Each radioisotope has a characteristic half-life, which is the time needed for half of a sample of radioisotope to undergo nuclear decay. Which quality is desirable for a radioisotope that is used for medical imaging of a specific organ
Physics
1 answer:
Natalija [7]3 years ago
5 0

Answer:

the distance/penetrative power

Explanation:

this is important since there are different electromagnetic waves used to observe different parts like some are used to detect tumours and growths and others fractures so to detect eg fractures the radiation must be highly penetrative to penetrate skin and flesh to reach the bones but it's half-life will

only matter when you look at environmental issues and stuff

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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov
Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from  point A to point B along

Voltage difference,ΔV =V₁ - V₂  

The work done

W = Q . ΔV

Given that  charge is moved from point A to point B along an equipotential surface.It means that voltage  difference is zero.

ΔV = 0

So

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0
3 years ago
In the sport of parasailing, a person is attached to a rope being pulled by a boat while hanging from a parachute-like sail. A r
IrinaK [193]

Answer:

570 N

Explanation:

Draw a free body diagram on the rider.  There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.

The rider is moving at constant speed, so acceleration is 0.

Sum of the forces in the x direction:

∑F = ma

F cos 30° - T cos 15° = 0

F = T cos 15° / cos 30°

Sum of the forces in the y direction:

∑F = ma

F sin 30° - W - T sin 15° = 0

W = F sin 30° - T sin 15°

Substituting:

W = (T cos 15° / cos 30°) sin 30° - T sin 15°

W = T cos 15° tan 30° - T sin 15°

W = T (cos 15° tan 30° - sin 15°)

Given T = 1900 N:

W = 1900 (cos 15° tan 30° - sin 15°)

W = 570 N

The rider weighs 570 N (which is about the same as 130 lb).

6 0
3 years ago
Which of the following is Not true about the noble gases
Alexandra [31]
Noble gases are not highly reactive
3 0
3 years ago
How much electrical energy is used by the river cooker oven rated 800 W ,230 V is switched on for 30 minutes
mariarad [96]

Answer:

E = 1440 kJ

Explanation:

It is given that,

Power of a cooker oven is 800 W

Voltage at which it is operated is 230 V

Time, t = 30 minutes = 1800 seconds

We need to find the electrical energy used by the cooker oven. The product of power and time is equal to the energy consumed. So,

E=P\times t\\\\E=800\ W\times 1800\ s\\\\E=1440000\ J\\\\\text{or}\\\\E=1440\ kJ

So, electrical energy of 1440 kJ is consumed by the cooker oven.

8 0
3 years ago
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