What is one way in which radioactive decay is used to benefit society?

2 answers:

7 0

<span>They are used to measure and map effluent and pollution discharges from factories and sewerage plants, and the movement of sand around harbours, rivers and bays. Radioactive materials used for such purposes have short half-lives and decay to background levels within days.</span>

Fudgin [204]2 years ago

4 0

Answer:

Sample Response: Radioisotopes are used in medicine as tracers to detect and diagnose medical problems. Radioisotopes are also used in radiation therapy to treat cancer.

On Edge

Explanation:

You might be interested in

Difference between duracell and interstate golf cart batteries

weeeeeb [17]

Duracell batteries are an example of an electrochemical cell that is powered between the reaction of Magnesium and Zinc, occurring in basic conditions (alkaline battery). This type of reaction has a precise output of 1.5 volts, and looks like this:

Zn + 2MnO2 ➡️ ZnO + Mn2O3

It’s not rechargeable.

Golf Cart Batteries are an example of an electrochemical cell that is powered by the reaction between Lead and Sulfuric Acid (Lead-Acid battery). This type of reaction occurs on larger scales than an alkaline battery, and thus can generate a variety of powers depending on how many instruments are present within the battery. The reaction looks like this:

PbO2 + Pb + 2H2SO4 ➡️ 2PbSO4 + H2O

This is a rechargeable cell, but is rather prone to discharging by the environment and surroundings of the battery.

3 0

2 years ago

Help! (Look in the pic)

hjlf

Answer:

turtle

Explanation:

5 0

2 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$