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:
η = 40 %
Explanation:
Given that
Qa ,Heat addition= 1000 J
Qr,Heat rejection= 600 J
Work done ,W= 400 J
We know that ,efficiency of a engine given as

Now by putting the values in the above equation ,then we get

η = 0.4
The efficiency in percentage is given as
η = 0.4 x 100 %
η = 40 %
Therefore the answer will be 40%.
Answer:
Explanation:
A general wave function is given by:

A: amplitude of the wave = 0.075m
k: wave number
w: angular frequency
a) You use the following expressions for the calculation of k, w, T and λ:



b) Hence, the wave function is:

c) for x=3m you have:

d) the speed of the medium:

you can see the velocity of the medium for example for x = 0:

Answer:
The main types of nucleons are protons and neutrons. A proton, as its name suggests, has a positive electric charge, and a neutron has a neutral electric charge (meaning that it has no charge). The two in the nucleus of the atom make a positive charge, since the neutron has no charge at all.
Explanation:
Answer:
the required minimum magnitude of the force F is 21 N
Explanation:
Given the data in the question,
m = 5 kg
width = 60 cm
height = 80 cm
Let force is F represent in the image below,
so when the block about to rotate normal shifted to edge of cube
mg(w/2) = Fh
F = mg(w/2) / h
we know that g = 9.8 m/s²
we substitute
F = (5 × 9.8 ( 60/2)) / 70
F = (5 × 9.8 × 30 ) / 70
F = 1470 / 70
F = 21 N
Therefore, the required minimum magnitude of the force F is 21 N