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

Answer 1
Answer:

(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.

HELP PLSSS I HAVE AN EXAM MONDAY AND I THINK THIS IS GONNA BE ON ITTTTA Fish Is At The Western Bank Of

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2021 edge

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Dad, if you are seeing this, then I have run away. Angie has been torturing me ever since you started dating her, and I have had enough of it. I have been trying to tell you, but you would never listen. Goodbye, and Angie, if you are seeing this, I posted this on here, so you will not be able to delete anything. Have a nice life

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u ok?

running away from problems only makes things worse. Your teaching urself to run. but I dont blame u

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Correct answer to your question is 477.8

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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

the bottom, and put it on

top, where it belongs:       (  4² · 3⁴ · 4² )  /  4³ · ( 3⁵    -- )

Multiply the 4²s on top:     (  4⁴ · 3⁴       )  /  ( 4³ · 3⁵ )

Now let me break this up.

Not changing anything, just

writing it in a different way:      ( 4⁴  /  4³ ) · ( 3⁴ / 3⁵ )

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

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There are a lot of other ways we could have manipulated and

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-- Remember that a number with a negative exponent belongs

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If the distance between the source of sound and observer is reduced to one half, then
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Answer:

Let the power delivered by the sound wave be 'P'

Intensity by definition is the power delivered per unit area.

i.e. I =

dA

dP

;

In spherical polar coordinate system dA = r

2

Ω

I =

r

2

Ω

dP

;

I ∝

r

2

1

So if distance increases 3 fold. The Intensity becomes

9

1

times the initial

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Answer:

718.02N is the answer to the question

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Explanation:

For this exercise let's use the first law of thermodynamics

             

          ΔE = Q + W

Where we will apply this expression to several cases

1) output Q = 500 J and does work of W = 300 J.

The two processes involve an energy output

         ΔE = - 500 - 300

         ΔE = -800 J

Therefore, the internal energy of the gas decreases by this amount,

2) enter Q = 500J and do work of W = 300 J

         Positive heat is added and does negative work

         ΔE = 500 - 300

         ΔE = 200 J

3) they leave Q = 500 J and do (work on the gas) W = 300J

negative heat and positive work come out

          ΔE = -500 + 300

          ΔE = -200 J

4) heat enters Q = 500 J, work on gas W = 300 J

both positive

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the crane takes 2.5s to raise the box 3.0 m. Calculate the power output of the crane

Answers

Answer:

Power = 1800 W

Explanation:

Power = work / time

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Power = 1800 W

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Answer:ur mom

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Answers

Answer:

Work is said to be done when a force applied to an object moves that object. We can calculate work by multiplying the force by the movement of the object.

Answer:

Work is done by a force on an object if (i) a force acts on the object and (ii) the object is displaced from its original position

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Explanation:

that is my answer thank me later

Answer:

oxygen difluoride a polar molecule


A 40kg girl and a 50kg boy are facing each other on a friction-free rollerskates. The girl pushes the boy, who moves away at a speed of 3 m/s. What is the girl's
speed?
Your answer

Answers

Answer:

Explanation:

The Law of Momentum Conservation, just like the Law of Thermodynamics about energy, says that momentum is neither created nor destroyed but is conserved, meaning it has to go somewhere. If the girl pushes the boy and they are both on friction free skates, then the girl will also react to the push. Momentum Conservation says

[tex](m_gv_g+m_bv_b)_b=(m_gv_g+m_bv_b)_a[/tex] In words this says that the mass times the velocity of the girl plus the mass times the velocity of the boy before the push has to equal the mass times velocity of the girl plus the mass times velocity of the boy after the push. Mathematically,

[tex](40.0*0+50.0*0)_b=(40.0v+50.0*-3)_a[/tex]

The left side of this is equal to 0. On the right, I made the velocity of the boy negative. We could have made it positive and it wouldn't have mattered. The sign will only be important to the result because if the sign of the girl's velocity is the same as the boy's, she is moving in the same direction as he is; if it's different, she is moving in the opposite direction.

0 = 40.0v - 150.0 and

-40.0v = -150.0 so

v = 3.75 This means that when she pushes the boy one way, mometum is conserved and she moves in the opposite direction and at a greater velocity (because her mass is less). Physics is a wonderful thing, isn't it!?

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