Answer: f = 927.55Hz
Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below
L = λ/4 where λ = wavelength.
speed of sound in air = v = 343m/s.
fundamental frequency of open closed tube = 315Hz
λ = 4L.
v = fλ
343 = 315 * 4L
343 = 1260 * L
L = 343/ 1260
L = 0.27m
In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.
The length of tube and wavelength are related by the formulae below
L = λ/4, λ=4L
λ = 4 * 0.27
λ = 1.087m.
v = fλ
1010 = f * 1.087
f = 1010/1.807
f = 927.55Hz
I only know about the Water tank which is the most accurate. You place your body in it, and weights are added I think. Somehow some measurements are gathered to get your body fat weight. Online calculators exist, as well as electronic waves that are sent int your body, the echo is recorded and analyzed.
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
(50 gal / 5 min) x (.0037854 m³/gal) x (1 min / 60 sec)
= (50 · 0.0037854 · 1) / (5 · 60) m³/sec
= 0.000631 m³/sec
Answer: 0°
Explanation:
Step 1: Squaring the given equation and simplifying it
Let θ be the angle between a and b.
Given: a+b=c
Squaring on both sides:
... (a+b) . (a+b) = c.c
> |a|² + |b|² + 2(a.b) = |c|²
> |a|² + |b|² + 2|a| |b| cos 0 = |c|²
a.b = |a| |b| cos 0]
We are also given;
|a+|b| = |c|
Squaring above equation
> |a|² + |b|² + 2|a| |b| = |c|²
Step 2: Comparing the equations:
Comparing eq( insert: small n)(1) and (2)
We get, cos 0 = 1
> 0 = 0°
Final answer: 0°
[Reminders: every letters in here has an arrow above on it]