HELP PLSSS I HAVE AN EXAM MONDAY AND I THINK THIS IS GONNA BE ON ITTTT
A fish is at the western bank of a river that is 64 m wide and has a current with a velocity of 0.90 m/s [S]. The fish swims directly across the river going due east. The fish can swim at a speed of 0.2 m/s.
a) How long does it take for the fish to get across the river
b) What is the resulting velocity of the fish
c) When the fish arrives on the opposite bank, how far is it from being at the point directly across from where it started?
The answers are here, please show work on how to get these
a) 3.2 x 10^2
b) 0.9 m/s (S 13 E)
c) 2.9 x 10^2
Answers
(a) 3.2 x 10²s
(b) 0.9 m/s (S 13 E)
(c) 2.9 x 10²m
Explanation:The sketch illustrating the scenario has been attached to this response.
As shown;
The fish swims due east with a velocity [tex]V_{x}[/tex] = 0.2m/s
The river current has a velocity [tex]V_{y}[/tex] 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 = [tex]\frac{d}{t}[/tex] -------------(i)
Where;
V = velocity of the fish = [tex]V_{x}[/tex] = 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 = [tex]\frac{d}{V}[/tex]
Substitute the values of d and V into the equation;
t = [tex]\frac{64m}{0.2m/s}[/tex]
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 [tex]V_{x}[/tex] and [tex]V_{y}[/tex], and direction is given by θ. i.e
The magnitude of the resulting vector is;
|V| = [tex]\sqrt{(V_x)^2 + (V_y)^2}[/tex]
|V| = [tex]\sqrt{(0.2)^2 + (0.9)^2}[/tex]
|V| = [tex]\sqrt{(0.04) + (0.81)}[/tex]
|V| = [tex]\sqrt{(0.85)}[/tex]
|V| = 0.92m/s
|V| ≅ 0.9m/s
The direction of the resulting vector θ and is given by;
tan θ = [tex]\frac{V_y}{V_x}[/tex]
tan θ = [tex]\frac{0.9}{0.2}[/tex]
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
This can also be written as S12.5°E
Approximating to the nearest whole number gives 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
Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.
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The principle of the law of conservation of momentum states that if any two objects undergo collision, then the total momentum of the objects before and after the collision will be the same if there is no external force acting on the colliding objects.
The Law of Conservation of Momentum has the equation which is:
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2.5 + 0 = 1.0 + 0.50 × v₂
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going to use here are:
-- Two multiply two numbers with the same base,
add their exponents. Like (A³) · (A²) = A⁵ .
But the bases have to be the same.
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A negative power on top means it belongs on the bottom.
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Like A⁻² means 1/A² . And 1/B⁻³ means B³ .
That's all you need in order to clean up the big fraction
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Original: ( 4⁻³ · 3⁴ · 4² ) / ( 3⁵ · 4⁻² )
Let's send the 4⁻³
to the bottom
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Not take that 4⁻² from
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Multiply the 4²s on top: ( 4⁴ · 3⁴ ) / ( 4³ · 3⁵ )
Now let me break this up.
Not changing anything, just
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Look at the first fraction: 4⁴ / 4³ .
Divide top and bottom by 4³ , and it becomes just 4 .
Now look at the second fraction: 3⁴ / 3⁵ .
Divide top and bottom by 3⁴ , and it becomes just 1/3 .
So in the end, we're left with just 4 / 3 .
And THAT is exactly equal to the original big messy fraction
in the question. It has exactly the same numerical value, but
you'd never know it when you see it, because it's a lot simpler.
There are a lot of other ways we could have manipulated and
massaged the original fraction, but the steps are the same:
-- Multiply numbers with the same base, by adding the exponents.
-- Remember that a number with a negative exponent belongs
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The wind blows a leaf at 37.9 m/s for 118 s left. How are does the leaf go in that period of time.
A) 3.11 m, left
B) 4470 m, left
C) 80.1 m, left
D) 0.321 m, left
Answers
Explanation:
distance = velocity * time
d = 37.9 * 118
distance = 4472.2
