Answer:
Explanation:
<h3><u>Given data:</u></h3>
Frequency = f = 200 Hz
Velocity = v = 400 m/s
<h3><u>Required:</u></h3>
Wavelength = λ = ?
<h3><u>Formula:</u></h3>
v = fλ
<h3><u>Solution:</u></h3>
Put the givens in the formula
400 = (200)λ
Divide 200 to both sides
400/200 = λ
2 m = λ
λ = 2 m
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Answer:
(a) 0.942 m
(b) 18.84 m/s
(c) 2366.3 m/s²
(d) 0.05 s
Explanation:
(a) In one revolution, it travels through one circumference, 2πr = 2 × 3.14 × 0.15 m = 0.942 m.
(b) Its frequency, f, is 1200 rev/min = 
rev/s = 20 rev/s.
Its angular frequency, ω = 2πf = 2π × 20 = 40π
The speed is given by
v = ωr = 40π × 0.15 = 6π = 18.84 m/s
(c) Its acceleration is given by, a = ω²r = (40π)² × 0.15 = 2366.3 m/s²
(d) The period is the inverse of the frequency because it is the time taken to complete one revolution.
T = 1/20 = 0.05 s
Answer:
Δy= 5,075 10⁻⁶ m
Explanation:
The expression that describes the interference phenomenon is
d sin θ = (m + ½) λ
As the observation is on a distant screen
tan θ = y / x
tan θ= sin θ/cos θ
As in ethanes I will experience the separation of the vines is small and the distance to the big screen
tan θ = sin θ
Let's replace
d y / x = (m + ½) λ
The width of a bright stripe at the difference in distance
y₁ = (m + ½) λ x / d
m = 1
y₁ = 3/2 λ x / d
Let's use m = 1, we look for the following interference,
m = 2
y₂ = (2+ ½) λ x / d
The distance to the screen is constant x₁ = x₂ = x₀
The width of the bright stripe is
Δy = λ x / d (5/2 -3/2)
Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)
Δy= 5,075 10⁻⁶ m
Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively
The rate of the change in temperature which is observed while moving upward in the earth's atmosphere with elevation. It can be positive, negative and zero when the temperature decreases, increases or is constant with the elevation respectively.
For the atmosphere, the drop in temperature of rising air that is unsaturated air is about 10 degrees C/1000 meters (5.5 °F per 1000 feet) altitude.
That means if a there occur a rise of 1000 m , then the temperature of that thing will decrease to 10 degrees. Every 10°C of temperature from the given temperature will decrease at every rise of 1000 m .Air that is not saturated will cool or heat at a rate of 10 degrees C/1000 meters as it rises or descends, respectively
To learn more about unsaturated air here
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