Answer:
s
Explanation:
It is given that, a physics problem you end up with m in the numerator and m/s in the denominator. It would mean that,

m is metre and s is second
or

so, the unit of the answer is <u>s</u>.
The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
<h3>What is a frequency?</h3>
The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.
Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.
Learn more about Frequency:
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c. The Moon is positioned directly between the earth and the sun is the statement that does not result in an eclipse.
Explanation:
- The Sun is completely blocked in a solar eclipse because the Moon passes between Earth and the Sun.
- it is just the right distance away from Earth, the Moon can fully blocks the Sun's light from Earth's perspective
Both, there are two different types of molecules to distinguish that
Answer:
a) K = 3 MeV b) K= 1.5 MeV
Explanation:
We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.
F = q v x B
We can also write this equation based on the modules of the vectors
F = qv B sin θ
With Newton's second law
F = ma
F = m v² / r
q v B = m v² / r
v = q B r / m
The kinetic energy is
K = ½ m v²
Substituting
K = ½ m (q B r/ m)²
K = ½ B² r² q² / m
K = (½ B² R²) q²/m
The amount in brackets does not change during the experiment
K = A q² / m
For the proton
K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J
With this data we can find the amount we call A
A = K m/q²
A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²
A = 3.13 10⁻²
With this value we can write the equation
K = 3.13 10⁻² q² / m
Alpha particle
m = 4 uma = 4 1.66 10⁻²⁷ kg
K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷
K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV
K = 3 MeV
Deuteron
K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷
K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)
K = 1.5 10⁶ eV
K= 1.5 MeV