Ask question

Ask question

All categories

xxTIMURxx [149]

3 years ago

11

Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa

ddles directly upstream (and moves upstream), whereas the other paddles directly downstream. With downstream as the positive direction, an observer on shore determines the velocities of the two canoes to be −1.2m/sand +2.9m/s, respectively. (a)What is the speed of the water relative to the shore? (b)What is the speed of each canoe relative to the water?

1 answer:

san4es73 [151]3 years ago

6 0

Answer:

a) speed of water relative to shore = 0.85 m/s

b) speed of each canoe relative to the water = 2.05 m/s

Explanation:

Let the speed of the canoeists be 'v'

and,

the speed of stream relative to shore be 'u'

therefore,

for the canoeist padding upstream

u - v = - 1.2 m/s .............(1)

and,

for the canoeist padding downstream

u + v = 2.9 m/s .............(2)

on adding equation (1) and (2), we get

u - v = - 1.2

+ u + v = 2.9

===========

2u + 0 = 1.7

or

u = 0.85 m/s

substituting u in equation 2 we get

0.85 + v = 2.9

or

v = 2.05 m/s

Hence,

a) speed of water relative to shore = 0.85 m/s

b) speed of each canoe relative to the water = 2.05 m/s

You might be interested in

Which statement about oceans is incorrect?

Hitman42 [59]

It is C because less than one percent of water is fresh water

7 0

3 years ago

An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

Svetllana [295]

Answer:

$v \approx 9.312\,\frac{m}{s}$

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

$K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}$

$K_{B} = K_{A} + U_{g,A}-U_{g,B} - W_{loss}$

$\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot s\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cdot s \cos \theta$

$\frac{1}{2}\cdot v^{2} = g\cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)$

$v = \sqrt{2\cdot g \cdot s \cdot (\sin \theta - \mu_{k}\cdot \cos \theta)}$

$v = \sqrt{2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (10\,m)\cdot (\sin 37^{\textdegree} - 0.2\cdot \cos 37^{\textdegree})}$

$v \approx 9.312\,\frac{m}{s}$

3 0

3 years ago

Which of the following cars have the most kinetic energy

faltersainse [42]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>

$\huge\boxed{OptionA}$

<h2>_____________________________________</h2><h2>DATA:</h2><h3>Blue Car: </h3>

mass = 4 kg

velocity = 5 m/s^2

<h3 /><h3>Orange truck:</h3>

Mass= 2kg

Velocity = 7m/s^2

<h3 /><h3>Grey Car:</h3>

mass = 6 kg

velocity = 4m/s^2

<h3 /><h3>Green Car:</h3>

Mass = 8 kg

Velocity = 3 m/s^2

<h2>_____________________________________</h2><h2>SOLUTION:</h2>

By the equation of kinetic energy,

K.E = $\frac{1}{2} mv^2$

Where,

K.E is kinetic energy

m is mass

v is velocity

<h2>_____________________________________</h2><h3>Kinetic energy of Blue car:</h3>

Directly substitute the variables in the equation,

K.E = $\frac{1}{2}x4x5^2$

Simplify the equation,

K.E = 50 J

<h2>_____________________________________</h2><h3>Kinetic Energy of Silver Car:</h3>

Directly substitude the variable in the equation,

K.E = $\frac{1}{2}x6x4^2$

Simplify the equation,

K.E = 48J

<h2>_____________________________________</h2><h3>Kinetic Energy of Green Car:</h3><h3 />

Substitute the variables in the equation,

K.E = $\frac{1}{2}x8x3^2$

Simplify the Equation,

K.E = 36J

<h2>_____________________________________</h2><h3>Kinetic Energy of Orange Truck:</h3><h3 />

Substitute the variable,

K.E = $\frac{1}{2}x 2x7^2$

Simplify the equation,

K.E = 49J

<h2>_____________________________________</h2>

As you can see that the highest value of kinetic energy is of Blue SUV thus it will be out answer.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 /><h3 /><h3 /><h3 /><h2 /><h2 />

6 0

3 years ago

A paper clip that has a mass of 1.5 grams is thrown into the air and initially has a kinetic energy

Vsevolod [243]

Answer:

$v = 4.2 \ m/s$

Explanation:

Given data:

Mass of the paper clip, $m = 1.5 \ g = 0.0015 \ kg$

Kinetic energy, $K = 0.013 \ \rm J$

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

$K = \frac{1}{2}mv^{2}$

$0.013 = 0.5 \times 0.0015 \times v^{2}$

$\Rightarrow \ v = 4.16 \ m/s$

$v = 4.2 \ m/s.$ (rounding to nearest tenth)

3 0

3 years ago

Your family decided to go to sea World in San Antonio this weekend. If it takes

Ne4ueva [31]

Answer:

63 mph

Explanation:

252/4 is 63mi

4 0

3 years ago

Read 2 more answers