Answer:
<h3> b. 1.18</h3>
Explanation:
The fundamental frequency in string is expressed as;
F1 = 1/2L√T/m .... 1
L is the length of the string
T is the tension
m is the mass per unit length
If the tension is increased by 40%, the new tension will be;
T2 = T + 40%T
T2 = T + 0.4T
T2 = 1.4T
The new fundamental frequency will be;
F2 = 1/2L√1.4T/m ..... 2
Divide 1 by 2;
F2/F = (1/2L√1.4T/m)/1/2L√T/m)+
F2/F = √1.4T/m ÷ √T/m
F2/F = √1.4T/√m ×√m/√T
F2/F = √1.4T/√T
F2/F = 1.18√T/√T
F2/F = 1.18
F2 = 1.18F
Hence the fundamental frequency of vibration changes by a factor of 1.18
Answer:
trigonometry (guessing)
Explanation:
ellipse: is the shape of an orbit : looks like an oval
periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth
parallax is triangulation. like how gps works. looking at a star one day and then looking at it again 6 months later, an astronomer can see a difference in the viewing angle for the star. With trigonometry, the different angles yield a distance. This technique works for stars within about 400 light years of earth
https://science.howstuffworks.com/question224.htm
By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars
1/r^2 rule states that the apparent brightness of a light source is proportional to the square of its distance.Jan 11, 2022
https://www.space.com/30417-parallax.html
alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years
blossoms.mit.edu
.
Color is what scientist use to determine the temperature of a star!
hope this helps!
Concept:
Frequency- It is defined as the number of oscillations occur in one second.
Its SI unit is Hertz (Hz)
Given: Produced sound vibrations is 18,500 cycles in 0.75 seconds
∵ In 0.75 second, produced sound has oscillations = 18,500 cycles
∴ In 1.0 second, produced sound has oscillations = (18,500 ÷ 0.75) Hz
The frequency of the sound will be ≈ 24,667 Hz
From the study of the given graph, only the animals (c) Cats, (b) Moths and (a) Bats can hear the produced sound because their upper audible frequency range is greater than 24,667 Hz.
Answer:
change waterslide according to question. and you are good to go. check photo for solve
