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lianna [129]
3 years ago
12

If an unknown element has a mass of 17 and contains 6 neutrons, how many protons does it have ?

Physics
1 answer:
lbvjy [14]3 years ago
6 0

The mass number is the total number of protons and neutrons within an atom and since we know that the unknown element has 6 neutrons, we can simply subtract the number of neutrons from the mass number to get the number of protons.

17 - 6 = 11

There are 11 protons in this unknown element.


Extra:

The number of protons (+) and electrons (-) are equal in a neutral atom so since you know that there are 11 protons you also know that there are 11 electrons. On the periodic table, the element with 11 electrons is Na or Sodium.


Hope this helps! :)

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Answer:

5. dispersion

6. 49.8°

Explanation:

5. Dispersion is the name given to the phenomenon of light of different wavelengths being bent differently. A rainbow is the result of light from a point source (the sun) being spread out by wavelength (color), a nice example of dispersion.

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6. n = 1.31 is the ratio of the sine of the angle of refraction to the sine of the angle of incidence (for light passing to a medium of n = 1). When the angle of refraction is 90°, the angle of incidence is the "critical angle." So, ...

sin(90°)/sin(critical) = 1.31

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Without:

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With safety:

  • The seatbelt holds tge passengers body back so they dont collide with anything
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b. add a magnetic metal core

Explanation:

a p e x test

8 0
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4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two
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Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=mV_{o}+MU_{o} (2)

p_{f}=(m+M)V_{f} (3)

m=1200 kg is the mass of the first car

V_{o}=20 m/s is the velocity of the first car, to the North

M=1400 kg is the mass of the second car

U_{o}=-22 m/s is the mass of the second car, to the South

V_{f} is the final velocity of both cars after the collision

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

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Finally:

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is there any choices?

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