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Readme [11.4K]
3 years ago
15

A man can swim with a speed of 5m/s in calm water. if this man swims crosses a specific river his speed is 3m/s. if he takes the

minimum time to cross the river find the speed of the flow of water in the river...
plz answer me with all the steps!
I'll give u points ..mark brainliest and follow u ​
Physics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

The speed of the flow of water in the river is 4 m/s.

Explanation:

Given that,

Speed of man = 5 m/s

If this man swims crosses a specific river his speed is 3 m/s.

If he takes the minimum time to cross the river

Let the speed of flow of water be v_{r}

We need to calculate the speed of the flow of water in the river

Using formula for velocity

v_{rs}^2=v_{r}^2+v_{s}^2

v_{r}^2=v_{rs}^2-v_{s}^2

Where, v_{s} = velocity of swimmer

v_{sr} = relative velocity

v_{r} = velocity of river

Put the value into the formula

v_{r}^2=5^2-3^2}

v_{r}=\sqrt{5^2-3^2}

v_{r}=4\ m/s

Hence, The speed of the flow of water in the river is 4 m/s.

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the answer is 568 yd^2

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3 years ago
A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un
Kamila [148]

Answer:

<em>a) 6738.27 J</em>

<em>b) 61.908 J</em>

<em>c)  </em>\frac{4492.18}{v_{car} ^{2} }

<em></em>

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

I = \frac{1}{2}mr^{2}

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

I =  \frac{1}{2}*11*1.1^{2} = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = Iw^{2} = 6.655 x 31.82^{2} = <em>6738.27 J</em>

<em></em>

b) second flywheel  has

radius = 2.8 m

mass = 16 kg

moment of inertia is

I = \frac{1}{2}mr^{2} =  \frac{1}{2}*16*2.8^{2} = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = Iw = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

(I_{1} +I_{2} )w

where the subscripts 1 and 2 indicates the values first and second  flywheels

(I_{1} +I_{2} )w = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = Iw^{2} = 6.655 x 3.05^{2} = <em>61.908 J</em>

<em></em>

<em></em>

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = \frac{1}{2}mv_{car} ^{2}

where m is the mass of the car

v_{car} is the velocity of the car

Equating the energy

2246.09 =  \frac{1}{2}mv_{car} ^{2}

making m the subject of the formula

mass of the car m = \frac{4492.18}{v_{car} ^{2} }

3 0
3 years ago
how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum
dangina [55]

Answer:

206.8965517 n

Explanation:

First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.

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2 years ago
What’s The Answer, To The Question In The Photo
Sedbober [7]

Answer:

The correct answer is the third option: The kinetic energy of the water molecules decreases.

Explanation:

Temperature is, in depth, a statistical value; kind of an average of the particles movement in any physical system (such as a glass filled with water). Kinetic energy, for sure, is the energy resulting from movement (technically depending on mass and velocity of a system; in other words, the faster something moves, the greater its kinetic energy.

Since temperature is related to the total average random movement in a system, and so is the kinetic energy (related to movement through velocity), as the thermometer measures <u>less temperature</u>, that would mean that the particles (in this case: water particles) are <u>moving slowly</u>, so that: the slower something moves, the lower its kinetic energy.

<u>In summary:</u> temperature tells about how fast are moving and colliding the particles within a system, and since it is <em>directly proportional</em> to the amount of movement, it can be related (also <em>directly proportional</em>) to the kinectic energy.

4 0
3 years ago
What is my displacement if i walked 200 meters west then turned around and walked 150 meters east
Alinara [238K]
You're walking in one direction, and then the exact opposite of that direction, so you simply have to subtract the two distances. 
200-150=50
You're 50 meters west of where you originally started.
You're west because 200 meters west is greater than 150 meters east. If the distance walked east was greater than the distance walked west, you would've been east of your starting position.
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