Answer: The correct answer is "the speed of the wave becomes four times".
Explanation:
The relation between the speed, frequency and the wavelength is as follows:
v=f\lambda
Here, v is the speed of the wave, f is the frequency and \lambda is the wavelength.
The speed of the sound wave is directly proportional to the frequency.
In the given problem, if the speed of the sound wave is increased four times then the speed of the sound becomes four times.
Therefore, the speed of the sound wave becomes four times.
A :-) for this question , we should apply
a = v - u by t
Given - u = 4.77 m/s
v = 23.5 m/s
t = 5.18 m/s
Solution -
a = v - u by t
a = 23.5 - 4.77
a = 28.27 m/s^2
.:. The acceleration is 28.27 m/s^2
It Would Take Aprroximately 1000 Years.
Answer:
Explanation:
331 m/s / 2.5e4 cyc/s = 0.01324 m ≈ 1.3 cm
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
