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Flauer [41]
3 years ago
12

What is a nanotoxin and what areas can they be found in?

Physics
1 answer:
slava [35]3 years ago
5 0
Nanotoxicology is the study of the toxicity of nanomaterials. Because of quantum size effects and large surface area to volume ratio, nanomaterials have unique properties compared with their larger counterparts that affect their toxicity.

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INSTI
max2010maxim [7]

Answer:

B

Explanation:

8 0
3 years ago
A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa
vlada-n [284]
Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3  meters per second
4 0
3 years ago
The first ionization energy of a hydrogen atom is 2.18 aj (attojoules). what is the frequency and wavelength, in nanometers, of
jasenka [17]

1) Frequency: 3.29\cdot 10^{15}Hz

the energy of the photon absorbed must be equal to the ionization enegy of the atom, which is

E=2.18 aJ=2.18\cdot 10^{-18} J

The energy of a photon is given by

E=hf

where h=6.63\cdot 10^{-34}Js is the Planck's constant. By using the energy written above and by re-arranging thsi formula, we can calculate the frequency of the photon:

f=\frac{E}{h}=\frac{2.18\cdot 10^{-18} J}{6.63\cdot 10^{-34} Js}=3.29\cdot 10^{15} Hz


2) Wavelength: 91.2 nm

The wavelength of the photon can be found from its frequency, by using the following relationship:

\lambda=\frac{c}{f}

where c=3\cdot 10^8 m/s is the speed of light and f is the frequency. Substituting the frequency, we find

\lambda=\frac{3\cdot 10^8 m/s}{3.29\cdot 10^{15}Hz}=9.12\cdot 10^{-8} m=91.2 nm

5 0
3 years ago
Wich usage of water uses the least amount in a year in the united states
Lapatulllka [165]

Live Stock because in 2010 live stock used 10,000 millions gallons of water per day but everything else was higher and irrigation is the highest with 115,000 million gallons per day.

6 0
3 years ago
Read 2 more answers
Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons
dybincka [34]
<span>The electric force is given by: 
 F = [ k*(q1)*(q2) ] / d^2 
 F = Electric force 
 k = Coulomb's constant 
 q1 = Charge of one proton 
 q2 = Charge of second proton 
 d = Distance between centers of mass 
 Values: 
 F = unknown 
 k = 8.98E 9 N-m^2/C^2 
 q1 = 1.6E-19 
 q2 = 1.6E-19 
 d = 1.0E-15 m 
 Insert values into F = [ k*(q1)*(q2) ] / d^2 
 F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2 
 F = </span>229.888 N
 answer
 the electric force of repulsion between nuclear protons is 229.888 N

3 0
3 years ago
Read 2 more answers
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