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Anestetic [448]
3 years ago
11

A man paddles a canoe in a long, straight section of a river. The canoe moves downstream with constant speed 5 m/s relative to t

he water. The river has a steady current of 1 m/s relative to the bank. The man's hat falls into the river. Five minutes later, he notices that his hat is missing and immediately turns the canoe around, paddling upriver with the same constant speed of 5 m/s relative to the water. How long does it take the man to row back upriver to reclaim his hat?
Physics
1 answer:
nikklg [1K]3 years ago
7 0

Answer:

Explanation:

We shall consider all movement with respect to water assuming that river is at rest or motionless .

speed of canoe = 5 m /s

in five minutes , distance between hat and canoe = 5 x 60 x 5

1500 m

This distance will be covered by man in return journey . In this case his speed is 5 m /s again considering river constant .

So this will be covered at 5 m /s

time taken = 1500 / 5 = 300 s

= 5 minutes .

so it will take 5 minutes to row back to reclaim his hat .

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The speed of sound at temperature of________in the air is 360m/s.​
Aleks04 [339]

Vo= 331+0.6T

360=331+0.6T

360-331=0.6T

29=0.6T

0.6T/29

T=6/290 so change it to simplest form and us formulas good luck

4 0
3 years ago
Read 2 more answers
Winds blowing toward the east are called easterlies.<br><br><br> True or False
eimsori [14]

Answer: False

Explanation:

Winds are named for the cardinal direction they blow from.  Hence, a wind that <em>"blows towards the east"</em>, logically should <u>come from the west </u>and is called a <em>"west wind"</em>.

In thise sense, one of the best examples of this type of wind are the <em>Westerlies</em>, which are are prevailing winds that blow from the west at midlatitudes and have the characteristic that are stronger during winter and weaker during summer.

Therefore, the statement is false.

6 0
3 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
3 years ago
Which of the following is true of high clouds?
Wewaii [24]
I’m pretty sure it’s B
3 0
2 years ago
A 1,000 kg car is driving on a 15 m high bridge at 5 m/s. What is the kinetic energy of the car?
yanalaym [24]

Answer:

KE=12,500J

Explanation:

The formula for kinetic energy is:

KE = \frac{1}{2}mv^2

We can plug in the given values into the equation:

KE = \frac{1}{2}*1000kg*(5m/s)^2

KE = 500kg*25m^2/s^2

KE=12,500J

7 0
2 years ago
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