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balandron [24]
3 years ago
7

A certain sea cow can paddle 2.0 m/s in still water. If she attempts to cross a river, from the south bank to the north with a c

urrent of 3.0 m/s flowing toward the east by paddling entirely perpendicularly to the flow of the river, in what direction will she be traveling relative to an observer on shore
Physics
1 answer:
aliina [53]3 years ago
4 0

Answer:

v = 3.6m / s ,   θ = 56º

Explanation:

This is a relative speed exercise, let's use the Pythagorean theorem

        v = √ (v₁² + v₂²)

where v₁ is the speed of the sea still water and v₂ the speed of the current

         

let's calculate

       v = √ (2² + 3²)

        v = 3.6m / s

to find the direction we use trigonometry

      tan θ = v₂ / v₁

       θ = tan⁻¹ (v₂ / v₁)

let's calculate

       θ = tan⁻¹ (3/2)

        θ = 56º

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I think that when a book hits the ground its potential energy converts into kinetic energy and then kinetic energy is transformed into sound and heat energy.

Explanation:

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Which of the following are characteristics of good experimental design? Check all that apply.
miv72 [106K]

Answer:

The correct answer will be-

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2. Tests only one variable at a time

3. Plans how to record data so they can be published

Explanation:

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Read 2 more answers
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

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