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dedylja [7]
3 years ago
12

Describe three uses of radioactive isotopes.

Physics
2 answers:
valina [46]3 years ago
8 0
1. Archaeological Dating. Carbon-14 is often used to find the age of a substance many years old.

2. In the use of x-rays and cat scans. X-Ray technicians often inject radioactive iodine in ones system to increase the contrast between soft tissue and bone on an x-ray image.

3. Smoke detectors: Americium is often used in smoke detectors because it is very sensitive to burning carbon dioxide.  

anzhelika [568]3 years ago
7 0

Answer:

Radioactive isotopes are used in medicine to detect medical problems and treat some diseases. They are used in agriculture to better understand the biological and chemical processes in plants. In geology, radioactive isotopes are used to determine the age of rocks and fossils. In archaeology, they are used to determine the age of ancient artifacts.

Explanation:

Just got it from the test.

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V125BC [204]

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Which type of light ray can produce identical but reversed
barxatty [35]

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Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

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            y ’= 450 sin 35

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4 0
3 years ago
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas
timurjin [86]

Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

          = 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

          = 0.135 J + 0.11220 J

          = 0.2472 J

          ≅ 0.247 J = 247 mJ

5 0
3 years ago
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