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18 days ago

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Alicia intends to swim to a point straight across a 100 m wide river with a current that flows at 1.2 m/s. She can swim 2.5 m/s

in still water. At what angle, measured from the upstream direction, must she swim upstream to achieve her goal

1 answer:

Sav [2.2K]18 days ago

4 0

Answer:

θ = 61.3°

Alicia must swim at an angle of 61.3°

Explanation:

Parameters given include:

Width of the river = 100 m

Alicia's speed in still water = 2.5 m/s

Speed of river's current = 1.2 m/s

The angle she needs to swim can be determined by combining the velocities, taking into account the current's influence.

Her swimming speed aimed against the current must offset the current's velocity;

2.5cosθ - 1.2 = 0

2.5cosθ = 1.2

cosθ = 1.2/2.5

θ = cosinverse(1.2/2.5)

θ = 61.3°

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Derive an algebraic equation for the vertical force that the bench exerts on the book at the lowest point of the circular path i

Keith_Richards [2263]

a)

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ii) 1.57 m/s

b)

i) Refer to the attached diagram

ii) Up

c) $N=mg+m\frac{v_b^2}{R}$

d) Greater than

Explanation:

The problem does not provide full details: consult the attachments for the complete text.

a)

The revolution period of the book equals the total duration needed for the book to make one full revolution.

By examining the graph, we can approximate the revolution period by calculating the time difference between two successive points of the book's motion that share the same shape.

We could use the time difference between two adjacent crests to estimate the period. The first crest is observed at t = 90 s, and the following crest appears at t = 210 s.

This results in the revolution period being

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ii)

The tangential speed of the book is computed as the ratio of the distance traveled over one revolution (i.e., the circumference of the wheel) to the revolution period.

Mathematically:

$v_b=\frac{2\pi R}{T}$

where

R represents the wheel radius

T = 120 s indicates the period

Based on the graph, the book reaches a maximum at x = +30 m and a minimum at x = -30 m, giving the diameter of the wheel as

d = +30 - (-30) = 60 m

This means the radius calculates to

R = d/2 = 30 m

So, the final speed is

$v_b=\frac{2\pi (30)}{120}=1.57 m/s$

b)

i) Please consult the attached free-body diagram for the book when at its lowest point.

Two forces act on the book at the lowest position:

- The weight of the book, represented as

$W=mg$

where m denotes the book's mass and g stands for gravitational acceleration. This force functions downward.

- The normal force the bench exerts on the book is represented by N. This force acts upward.

ii)

While at its lowest position, the book maintains a horizontal motion at constant speed.

Nevertheless, the book is undergoing acceleration. Acceleration is defined as the rate of velocity change, which is vectorial, having both speed and direction. While the speed remains unchanged, the direction changes (upward), indicating the book has upward net acceleration.

According to Newton's second law, the net vertical force acting on the book corresponds with the vertical acceleration:

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c)

As discussed in part b), there are two forces influencing the book at the lowest point:

- The weight, $W=mg$ , directed downward

- The normal force from the bench, N, directed upward

Given that the book is in uniform circular motion, the net force must match the centripetal force $m\frac{v_b^2}{R}$ , leading us to the equation:

$N-mg=m\frac{v_b^2}{R}$

where

$v_b$ represents the speed of the book

R stands for the radius of the circular path.

We derive an expression for the normal force:

$N=mg+m\frac{v_b^2}{R}$

d)

As per the discussions in parts c) and d):

- The normal force acting on the book at its lowest point becomes

$N=mg+m\frac{v_b^2}{R}$

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$W=mg$

Upon comparing these two equations, we conclude:

$N>W$

Thus, it is evident that the normal force exerted by the bench exceeds the weight of the book.

4 0

1 month ago