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Paladinen [302]
3 years ago
6

Are Carbon and oxygen are both metalloids?

Physics
1 answer:
Marizza181 [45]3 years ago
7 0
Just carbon is a metalloid not oxygen
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A system has two possible energy states, E0 and E1 (with E1 > E0). If the difference between these energy states grows (E0 an
Ugo [173]

Answer:

The number of particles in state E0 over the number of particles in state E1  will reduce

Explanation:

E0 represents the ground level state when all the particles have same energy level.

E1 represents excited state in which only a few particle reaches

E0 and E1 get further apart  means that the energy difference between the two level increases.

Thus, the number of particles in state E0 over the number of particles in state E1  will reduce.

5 0
3 years ago
Point charge μC is located at x =, y = , point charge is located at x = 0m. What are (a)the magnitude and (b)direction of the to
pishuonlain [190]

Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0

Explanation:

  • a) First of all find the distance between the two charges;
  • x = 0, y = 0.30  and x = 0.40 m, y = 0
  • r = √( 0.4² + 0.3²)
  • = 0.5m

hence, the force F = 2Kq1q2cosθ /r²...............equation 1

but cosθ = y/r = 0.3/0.5

cosθ = 0.6

plugging back to equation 1;

F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2

F = 540 x 10^-3

Magnitude of Force = 0.54N

b) Direction is at angle 90

6 0
3 years ago
two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f
Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

E = 1.44 \times 10^7 N/C

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

E = 2\frac{kq}{r^2}

now we have

q = 2\times 10^{-6} C

r = 5 cm = 0.05 m

now we have

E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}

E = 1.44 \times 10^7 N/C

6 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
Describe a situation when you might travel at a high velocity, but with low acceleration
Alisiya [41]

Reading a book in your warm, comfy seat ... in Row-27 of a
passenger airliner cruising at 450 miles per hour.
 
5 0
3 years ago
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