Change in velocity of larger moose: (1/3)v - v = -(2/3)v
<span>change in velocity of small moose: (1/3)v - (-v) = (4/3)v </span>
<span>- (change in velocity of larger moose)/(change in velocity of smaller moose) = 2
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Answer:
15.1°
Explanation:
The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:
(1)
where
is the initial vertical velocity
g = 9.8 m/s^2 is the gravitational acceleration
t is the time
Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

Substituting t into (1) we find the final vertical velocity

where the negative sign means that the velocity is downward.
Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

Answer:
5) 4.00, 2.00, 1.0
Explanation:
wave equation is given as;
F₀ = V / λ
Where;
F₀ is the fundamental frequency = first harmonic
Length of the string for first harmonic is given as;
L₀ = (¹/₂) λ
λ = 2 L₀
when L₀ = 1
λ = 2 x 1 = 2m
when L₀ = 2m
λ = 2 x 2 = 4m
For First harmonic, the wavelength is 2m, 4m
For second harmonic;
L₁ = (²/₂)λ
L₁ = λ
When L₁ = 1
λ = 1 m
when L₁ = 2
λ = 2 m
For second harmonic, the wavelength is 1m, 2m
Thus, the wavelength that could represent harmonics present on both strings is 4m, 2m, 1 m
Answer:
i am answering the same question 3rd time
please find the answer in the images attached.