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Elis [28]
3 years ago
13

In order to control bacterial infections and/or food poisoning, the FDA (Food and Drug Administration) has approved the use of n

uclear energy to control pathogens and extend shelf life of refrigerated and frozen uncooked meat. Which nuclear energy application would be most useful for this situation?
Physics
2 answers:
ololo11 [35]3 years ago
8 0

Answer: Irradiation

Explanation: Irradiation is the process by which the food or the package is first exposed to the ionizing radiation.

The radiations like gamma rays, x-rays or electron beam are exposed to the food which increases the shelf life of the food by killing the harmful bacteria in it or by preventing the growth of harmful bacteria on the food.

It basically aids in killing the bacteria that causes the food spoilage.

Lelechka [254]3 years ago
4 0

The answer is food irradiation. This involves the brief exposure of food to gamma rays or X-rays to kill pathogens that may contribute to food spoilage. This increases the shelf-life of the food. Gamma rays and X-rays emanate from nuclear decay of radioactive materials such as uranium..

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You drop a ball from a height of 32 meters how much time passes before the ball hits the ground
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31 seconds

Explanation:

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What is Restoring force for large waves is A. Wind, B. Capillary Action, C. Diffusion D.Friction, or E. Gravity
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We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes
Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

1.56421\times 10^{-6}\ T

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = \frac{d}{2}=\frac{7}{2}=3.5\ cm

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

c = Speed of light = 3\times 10^8\ m/s

The intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2

5% of this energy goes to the visible light so the intensity is

I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

u=\frac{1}{2}\epsilon_0E^2

Energy density is also given by

\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T

The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

7 0
3 years ago
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