Answer:
The tension in the string is 
.
Explanation:
For a string with tension 
and linear density 
carrying a transverse wave at speed 
it is true that
solving for we get:
Now, the transverse wave covers the distance of 7.4mm in 0.88s, which means it's speed is
And it's linear density (mass per unit length) is
Therefore, the tension in the cord is
or in micro newtons
Answer:
compared to the incident angle, the refracted angle is 45.56⁰
Explanation:
From Snell's law;
n₁sin(I) = n₂sin(r)
Where;
n₁ is the refractive index of light in medium 1 = 1.2
n₂ is the refractive index of light in medium 2 = 1.4
I is the incident angle
r is the refractive angle
I = 56.439⁰
Applying snell's law
Therefore, compared to the incident angle, the refracted angle is 45.56⁰
If an object is changing it is called velocity - whether by a constant amount or a varying
Answer:
a) f=0.1 Hz ; b) T=10s
c)λ= 36m
d)v=3.6m/s
e)amplitude, cannot be determined
Explanation:
Complete question is:
Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.
Given:
number of wave crests 'n'= 5
pass in a time't' 54.0s
distance between two successive crests 'd'= 36m
a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have
f=n/t
f= 5/ 54 => 0.1Hz
b)The time period of wave 'T' is the reciprocal of the frequency
therefore,
T=1/f
T=1/0.1
T=10 sec.
c)wavelength'λ' is the distance between two successive crests i.e 36m
Therefore, λ= 36m
d) speed of the wave 'v' can be determined by the product of frequency and wavelength
v= fλ => 0.1 x 36
v=3.6m/s
e) For amplitude, no data is given in this question. So, it cannot be determined.
