Answer:
1a) 857143 m
1b) 414 m
2a)
2b)
3) the medium of air has a wavelength of 0.334 m, the medium of water has a wavelength of 1.493 m, and the medium of 5.130 m.
Explanation:
Question 1a)
Given the velocity/speed, and frequency of the wave, the length can be calculated using these two quantites.
[ λ = v / f ] wavelength = <u>v</u>elocity of the wave / <u>f</u>requency of the wave in Hz.
Since 3 × 10^8 × ms^-1 is the velocity, and 350Hz is the frequency.
Anything to the negative power is reciprocated. i.e ms^-1 = m/s.
The wavelength is 300000000m/350Hz = 857142.8571428..... m ≈ 857143 m
Question 1b) Given that the frequency of the second wave in water is 1% of the first wave, and the speed of the second wave is 1450ms^-1
Therefore the second wave has a frequency of 1% of 3.5 = 350/100 Hz = 3.5 Hz
The wavelength is found using the same
formula: wavelength = 1450m/3.5Hz = 414.2857142857.... m ≈ 414 m
Question 2a)
Question 2b)
Question 3) Remember, the speed of sound of the medium = frequency of the medium × wavelength of the medium.
Therefore the wavelength of the medium = speed of sound of the medium / frequency of the medium. This has a similar correlation to the wavelength formula. We are given that all these mediums have a frequency of 1KHz = 1000Hz, where So the wavelength of each medium =
Question 4)
(1) The linear acceleration of the yoyo is 3.21 m/s².
(2) The angular acceleration of the yoyo is 80.25 rad/s²
(3) The weight of the yoyo is 1.47 N
(4) The tension in the rope is 1.47 N.
(5) The angular speed of the yoyo is 71.385 rad/s.
<h3> Linear acceleration of the yoyo</h3>
The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.
∑τ = Iα
rT - Rf = Iα
where;
- I is moment of inertia
- α is angular acceleration
- T is tension in the rope
- r is inner radius
- R is outer radius
- f is frictional force
rT - Rf = Iα ----- (1)
T - f = Ma -------- (2)
a = Rα
where;
- a is the linear acceleration of the yoyo
Torque equation for frictional force;
solve (1) and (2)
since the yoyo is pulled in vertical direction, T = mg 
<h3>Angular acceleration of the yoyo</h3>
α = a/R
α = 3.21/0.04
α = 80.25 rad/s²
<h3>Weight of the yoyo</h3>
W = mg
W = 0.15 x 9.8 = 1.47 N
<h3>Tension in the rope </h3>
T = mg = 1.47 N
<h3>Angular speed of the yoyo </h3>
v² = u² + 2as
v² = 0 + 2(3.21)(1.27)
v² = 8.1534
v = √8.1534
v = 2.855 m/s
ω = v/R
ω = 2.855/0.04
ω = 71.385 rad/s
Learn more about angular speed here: brainly.com/question/6860269
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What’s the question? I may be able to help
Answer:
t = 2.13 10-10 s
, d = 6.39 cm
Explanation:
For this exercise we use the definition of refractive index
n = c / v
Where n is the refraction index, c the speed of light and v the speed in the material medium.
The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is
v = c / n
As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships
v = d / t
t = d / v
t = d n / c
Let's look for the times on each sheet
Ice
t₁ = 1.4 10⁻² 1.31 / 3 10⁸
t₁ = 0.6113 10⁻¹⁰ s
Crown glass (BK7)
t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸
t₂ = 1.52 10⁻¹⁰ s
Time is a scalar therefore it is additive
t = t₁ + t₂
t = (0.6113 + 1.52) 10⁻¹⁰
t = 2.13 10-10 s
The distance traveled by this time in a vacuum would be
d = c t
d = 3 10⁸ 2.13 10⁻¹⁰
d = 6.39 10⁻² m
d = 6.39 cm
