Two violinists are trying to play in tune. However, whenever they play their A string at the same time they hear a beat frequenc
1 answer:
Answer:
The possible frequencies for the A string of the other violinist is 457 Hz and 467 Hz.
(3) and (4) is correct option.
Explanation:
Given that,
Beat frequency f = 5.0 Hz
Frequency f'= 462 Hz
We need to calculate the possible frequencies for the A string of the other violinist
Using formula of frequency
...(I)
...(II)
Where, f= beat frequency
f₁ = frequency
Put the value in both equations
Hence, The possible frequencies for the A string of the other violinist is 467 Hz and 457 Hz.
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<h2>Answer:</h2>
(a) 3.2 x 10²s
(b) 0.9 m/s (S 13 E)
(c) 2.9 x 10²m
<h2>Explanation:</h2>The sketch illustrating the scenario has been attached to this response.
As shown;
The fish swims due east with a velocity = 0.2m/s
The river current has a velocity due South = 0.9m/s
The resultant of the velocity is V
The width of the river is x = 64m
(a) To calculate how long it took the fish to get across the river, we know that velocity is the rate of change in distance, therefore we can use the relation;
V = -------------(i)
Where;
V = velocity of the fish = = 0.2m/s
d = distance from the start to the end = width of the river = x = 64m
t = time taken to move for that distance
Make t subject of the formula in equation (i);
t =
Substitute the values of d and V into the equation;
t =
t = 320 s
t = 3.20 x 10²s
Therefore, the time taken for the fish to get across the river is 3.20 x 10²s
(b) The resulting vector of the fish is V whose magnitude is the algebraic sum of vectors and
, and direction is given by θ. i.e
<em>The magnitude of the resulting vector is;</em>
|V| =
|V| =
|V| =
|V| =
|V| = 0.92m/s
|V| ≅ 0.9m/s
<em>The direction of the resulting vector θ and is given by;</em>
tan θ =
tan θ =
tan θ = 4.5
θ = tan⁻¹ ( 4.5)
θ = 77.47° South of East.
θ ≈ 77.5° South of East.
Subtracting θ = 77.5° from 90° gives its value East of South
i.e
90 - 77.5 = 12.5° East of South
<em>This can also be written as S12.5°E</em>
<em>Approximating to the nearest whole number gives </em>S 13 E
Therefore, the resulting velocity of the fish is 0.9m/s in the direction S13°E
(c) When the fish arrives on the opposite bank, its distance from being at the point directly across from where it started is the product of the velocity of the river current and the time taken by the fish to get across the river. This point is equivalent to k as shown in the diagram.
Therefore;
distance = velocity of river current x time taken
distance = 0.9m/s x 3.20 x 10²s
distance = 2.88 x 10²m
distance ≅ 2.9 x 10²m
<em>Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.</em>
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