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.
6 3/7 * 1 5/9
45/7 * 14/9
630/63
10
Answer:
Star A is closer than Star B
Explanation:
As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle
Here we can say
so we have
so here we have
angle subtended by Star A = 1 arc sec
angle subtended by star B = 0.75 arc sec
now we have
distance for star A is given as
distance of star B is given as
So star A is closer than star B
I believe all of these would be known as specific phobias.
