Hihi!
The correct answer is B) <span>neutron keep protons apart so they don’t repel
each other! </span><span>The </span>neutron<span> also adds mass to the </span>atom<span>!
</span>
I hope I helped!
-Jailbaitasmr
Answer:
a) The minimum thickness of the oil slick at the spot is 313 nm
b) the minimum thickness be now will be 125 nm
Explanation:
Given the data in the question;
a) The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot?
t
= λ/2n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.20
we substitute
t
= 750 / 2(1.20)
t
= 750 / 2.4
t
= 312.5 ≈ 313 nm
Therefore, The minimum thickness of the oil slick at the spot is 313 nm
b)
Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now?
minimum thickness of the oil slick at the spot will be;
t
= λ/4n
given that; wavelength λ = 750 nm and index of refraction of the oil n = 1.50
we substitute
t
= 750 / 4(1.50)
t
= 750 / 6
t
= 125 nm
Therefore, the minimum thickness be now will be 125 nm
Energy can not be created or destroyed but can change from one form to another.
example: as a roller coaster cart loses height the more speed it gains, the potential energy is transferred into kenetic energy
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
The frequency of the wave is 6800 Hz
<u>Explanation:</u>
Given:
Wave number, n = 20
Speed of light, v = 340 m/s
Frequency, f = ?
we know:
wave number = 

Therefore, the frequency of the wave is 6800 Hz