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:
B. 7.07 m/s
Explanation:
The velocity of the stone when it leaves the circular path is its tangential velocity,
, which is given by

where
is the angular speed and
is the radius of the circular path.
is given by

where
is the frequency of revolution.
Thus

Using values from the question,

<em>Note the conversion of 75 cm to 0.75 m</em>

Answer:
Final velocity of electron,
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A, 
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
........(1)
a is the acceleration, 
We know that electric force, F = qE

Use above equation in equation (1) as:


v = 647302.09 m/s
or

So, the final velocity of the electron when it reaches point B is
. Hence, this is the required solution.
b. 460.8 m/s
Explanation:
The relationship between the speed of the wave along the string, the length of the string and the frequency of the note is

where v is the speed of the wave, L is the length of the string and f is the frequency. Re-arranging the equation and substituting the data of the problem (L=0.90 m and f=256 Hz), we can find v:

c. 18,000 m
Explanation:
The relationship between speed of the wave, distance travelled and time taken is

where
v = 6,000 m/s is the speed of the wave
d = ? is the distance travelled
t = 3 s is the time taken
Re-arranging the formula and substituting the numbers into it, we find:
