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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?

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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saw5 [17]

The potential energy of the block is A) 490 J

Explanation:

The potential energy of an object is the energy possessed by the object due to its position in the gravitational field.

It is calculated as follows:

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object above the ground

For the block in this problem, we have:

m = 10 kg

$g=9.8 m/s^2$

h = 5 m

Therefore, its potential energy is:

$PE=(10)(9.8)(5)=490 J$

Learn more about potential energy:

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3 years ago

In which case does viscosity play a dominant role? Case A: a typical bacterium (size ~ 1 mm1 mm and velocity ~ 20 mm/s20 mm/s) i

My name is Ann [436]

Answer:

Case A

Explanation:

given,

size of bacteria = 1 mm x 1 mm

velocity = 20 mm/s

size of the swimmer = 1.5 m x 1.5 m

velocity of swimmer = 3 m/s

Viscous force

$F = \eta A \dfrac{dv}{dx}$

for the bacteria

$F = \eta \times 10^{-6}\times 20\times 10^{-3}$

$F =2\times 10^{-8} \eta\ N$

for the swimmer

$F = \eta \times 1.5^2\times 3$

$F =6.75 \eta\ N$

from the above force calculation

In case B inertial force that represent mass is more than the inertial force in case of bacteria.

Viscous force is dominant in case of bacteria.

So, In Case A viscous force will be dominant.

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3 years ago

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Paraphin [41]

Answer:

Simple awnser Do it yourself I really would help but I have no clue! Sorry

Explanation:

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3 years ago

Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc

snow_lady [41]

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force $F_g$ = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

5 0

3 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :

$d=\frac{1}{2}a{t}^{2}$ (10)

We already know the value of $d$ and calculated $a$ , we have to find $t$ :

$t=\sqrt{\frac{2d}{a}}$ (11)

$t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}$ (12)

$t=0.50s$ (13) >>>This is the time it takes to the block to go from the initial velocity $V_{o}$ to its final velocity $V_{f}$

If the acceleration is the variation of the velocity in time, we can use the following equation to find $V_{f}$ :

$V_{f}-V_{o}=a.t$ (13)

If $V_{o}=0$

$V_{f}=a.t$ (14)

$V_{f}=(5.88m/{s}^{2})(0.50s)$ (15)

Finally we get the value of the Final Velocity of the block:

$V_{f}=2.96m/s$

6 0

3 years ago