Answer:
4 hrs
Explanation:
625 = 5000 (1/2)^n
625/5000 = 1/2^n
n = 3 half lives
3 half lives = 12 hours 1 half life = 4 hrs
This question is incomplete; here is the complete question:
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32.4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate?
A. The wave has traveled 32.4 cm in 3 seconds.
B. The wave has traveled 32.4 cm in 9 seconds.
C. The wave has traveled 97.2 cm in 3 seconds.
D. The wave has traveled 97.2 cm in 1 second.
The answer to this question is D. The wave has traveled 97.2 cm in 1 second.
Explanation:
The frequency of a wave, which is in this case 3 hertz, represents the number of waves that go through a point during 1 second. According to this, if the frequency of the wave is 3 hertz this means in 1 second there were 3 waves. Moreover, if you multiply the wavelength (32.4cm) by the frequency (3) you will know the distance the wave traveled in 1 second: 32.4 x 3 = 97.2 cm. This makes option D the correct one as the distance in 1 second was 97.2 cm.
Answer:
(1) Hooke's law
(2) a) Extension is directly proportional to the applied load
b) The starting point of the graph is the origin (0, 0) or absence of load, no extension
Explanation:
(1) The law obeyed by the spring is known as Hooke's law which states that the extension or compression, x, of a spring proportional to the applied force, F
F = -k × x
Where;
k = The spring constant
(2) Given that the law mathematically is F = -k × x
The two features of the graph that show that the law is obeyed are;
a) The extension increases as the load is increased
b) The extension is zero when the there is no applied load.
Answer:
None of the wavelength is in the visible range
Explanation:
Constructive interference of the reflected waves for different wavelengths can be estimated using:
λ
= 2nd/m
where m is 1,2,3, ...
Therefore:
m=1, λ
= 750 nm
m=2, λ = 750/2 = 375 nm
The limits of eye's sensitivity is between 430 nm and 690 nm. Beyond this range, the eye's sensitivity drops to approximately 1% of its maximum value.
Answer:
x = 259 Hz
Explanation:
given,
frequency of one tuning fork = 250 Hz
frequency of another tuning fork = 266 Hz
when a tuning fork is sounded together beat frequency heard = 9
let x be the frequency of unknown
x - 250 = 9 Hz..............(1)
x = 259 Hz
when a another tuning fork is sounded together beat frequency heard = 7
266 - x = 7 Hz..............(2)
x = 259 Hz
now, on solving both the equation the frequency comes out to be 259 Hz.
so, The frequency of the tuning fork is equal to 259 Hz
