The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.
<h3>What is cutoff frequency?</h3>
The work function is related to the frequency as
W0 = h x fo
where, fo = cutoff frequency and h is the Planck's constant
Given is the work function for magnesium is 3.70 eV.
fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴
fo = 8.93 x 10¹⁴ Hz.
Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.
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Answer:
3.28 cm
Explanation:
To solve this problem, you need to know that a magnetic field B perpendicular to the movement of a proton that moves at a velocity v will cause a Force F experimented by the particle that is orthogonal to both the velocity and the magnetic Field. When a particle experiments a Force orthogonal to its velocity, the path it will follow will be circular. The radius of said circle can be calculated using the expression:
r = 
Where m is the mass of the particle, v is its velocity, q is its charge and B is the magnitude of the magnetic field.
The mass and charge of a proton are:
m = 1.67 * 10^-27 kg
q = 1.6 * 10^-19 C
So, we get that the radius r will be:
r = 
= 0.0328 m, or 3.28 cm.
Answer:
s = 30330.7 m = 30.33 km
Explanation:
First we need to calculate the speed of sound at the given temperature. For this purpose we use the following formula:
v = v₀√[T/273 k]
where,
v = speed of sound at given temperature = ?
v₀ = speed of sound at 0°C = 331 m/s
T = Given Temperature = 10°C + 273 = 283 k
Therefore,
v = (331 m/s)√[283 k/273 k]
v = 337 m/s
Now, we use the following formula to calculate the distance traveled by sound:
s = vt
where,
s = distance traveled = ?
t = time taken = 90 s
Therefore,
s = (337 m/s)(90 s)
<u>s = 30330.7 m = 30.33 km</u>
Work = Force x Distance
Assuming that this work is being done parallel to the displacement that is, but under that assumption:
W = (50)(10)
W = 500 J
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