Answer: T = 472.71 N
Explanation: The wire vibrates thus making sound waves in the tube.
The frequency of sound wave on the string equals frequency of sound wave in the tube.
L= Length of wire = 26cm = 0.26m
u=linear density of wire = 20g/m = 0.02kg/m
Length of open close tube = 86cm = 0.86m
Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as
L = 3λ/4
0.86 = 3λ/4
3λ = 4 * 0.86
3λ = 3.44
λ = 3.44/3 = 1.15m.
Speed of sound in the tube = 340 m/s
Hence to get frequency of sound, we use the formulae below.
v = fλ
340 = f * 1.15
f = 340/ 1.15
f = 295.65Hz.
f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.
The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as
L = λ/2
0.26 = λ/2
λ = 0.52m
The speed of sound in string is given as v = fλ
Where λ = 0.52m f = 295.65 Hz
v = 295.65 * 0.52
v = 153.738 m/s.
The velocity of sound in the string is related to tension, linear density and tension is given below as
v = √(T/u)
153.738 = √T/ 0.02
By squaring both sides
153.738² = T / 0.02
T = 153.738² * 0.02
T = 23,635.372 * 0.02
T= 472.71 N
Answer:
τ ≈ 0.90 N•m
F =
Explanation:
I = ½mR² = ½(10)0.5² = 1.25 kg•m²
α = ω²/2θ = 3.0² / 4π = 0.716... rad/s²
τ = Iα = 1.25(0.716) = 0.8952... ≈ 0.90 N•m
τ = FR
Now we have the unanswered question of reference frame.
80° from what?
If it's 80° from the radial
F = τ/Rsinθ = 0.90/0.5sin80 = 1.818... ≈ 1.8 N
If it's If it's 80° from the tangential
F = τ/Rcosθ = 0.90/0.5cos80 = 10.311... ≈ 10 N
There are an infinite number of other potential solutions
Answer:
Star A would be brighter than Star B
Explanation:
The apparent brightness of a star as perceived on Earth is dependent on its temperature, size, luminosity and distance from the Earth. Apparent brightness is the visible brightness to the eye at the surface of the Earth, while luminosity is the true brightness at the surface of the star.
A hotter star will radiate more energy per second per meter square of surface area. A larger star will have a greater surface area for radiation of energy, thus increasing the luminosity. For two identical stars, the difference in apparent brightness will be dependent on their distances from Earth.
Brightness and distance from earth have an inverse square relationship.
∝
Assuming the star is a point source of radiation, as distance from the source is increased, the radiation is distributed over a surface proportional to the distance form the source. As distance is further increased, the radiation is distributed over a larger surface area reducing the effective luminosity.
If one star (Star B) is twice as far from the earth as the first (Star A), the brightness of Star B will be
of Star A.
Thus, Star B will appear to be a quarter of the brightness of Star A. Or, Star A will appear to be 4 times as bright as Star B.
Answer:
Evaporation increases with an increase in the surface area
If the surface area is increased, then the amount is of liquid that is exposed to air is larger. More molecules can escape with a wider surface area.
Explanation: