Frequency = 1 / (period)
Frequency = 1 / (10 seconds) = (1/10) ( / second) = 0.1 per second = <em>0.1 Hz</em>.
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.
Learn more about cutoff frequency.
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Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
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Answer:
v_average = 15 m / s
Explanation:
The average speed can be found in two ways,
* taking the distance traveled and divide it by the time spent
* taking the velocities in each time interval and then finding the weighted average by the time fraction
v_average = 1 / t_total ∑
vi ti
Let's apply this last equation
Total time is
t = t₁ + t₂
t = 10 + 10 = 20 min
v_average = 10/20 10 + 10/20 20
v_average = 10/2 + 20/2
v_average = 15 m / s