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2 years ago

What is gama rays an it's uses

Physics

1 answer:

pav-90 [236]2 years ago

5 0

Answer:

GAMMA RAYS:

A photon emitted spontaneously by a radioactive substance also : a photon of higher energy than that of an X-ray.

USES OF GAMMA RAYS:

Gamma rays are used in medicine (radiotherapy), industry (sterilization and disinfection), and the nuclear industry. Shielding against gamma rays is essential because they can cause diseases to skin or blood, eye disorders, and cancers.

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The total amount of fresh water on earth is estimated to be 3.73 x 10^8 km^3. What is this volume in cubic meters? in cubic feet

Burka [1]

One km^3 is 1,000,000,000 m^3=10^9 m^3 hence 3.73 10^8 km^3 is 3.73 10^17 m^3

One meter is 3.28084 feet hence 1 m^3 is (3.28084)^3 feet

Thus 3.73 10^8 km^3 is 3.73*35.315 10^17 = 132 cubic feet

5 0

2 years ago

Read 2 more answers

Which property best makes radio waves safe for diagnosing illnesses through magnetic resonance imaging?

aleksandrvk [35]

Answer:

It is used in MRI because it does not damage cells

Radio waves are used for space research because they have very long wavelengths

Explanation:

Many parts of the electromagnetic spectrum are applied in clinical diagnosis and treatment of illnesses. However, these highly ionizing radiation damage cells and its dosage must be carefully managed to avoid creating radiation related health problems for the patients.

Radio waves can be used in MRI without issues because the energy of the radiation is not sufficient to cause damage to cells but is sufficient to provide images for the sake of medical diagnosis.

Secondly, radio waves have long wavelength. This property is suitable for long range

communication. Hence it can be used in space research

4 0

2 years ago

Read 2 more answers

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .