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3 years ago

What is the flow of energy from the falling water to the steam?

Physics

1 answer:

kenny6666 [7]3 years ago

3 0

Answer:

The flow of energy from falling water to the steam is;

a) Mechanical → Mechanical → Electrical → Thermal → Thermal

Explanation:

1) Mechanical → Mechanical

The water in the pipe before it falls possesses potential energy which it converts into kinetic energy as it falls from height

2) Mechanical → Mechanical

The water falling from the pipe stream unto the turbine wheel transfers its kinetic (mechanical) energy due to its motion on to the turbine wheel to give the wheel rotational motion

3) Mechanical → Electrical

The kinetic (mechanical) energy from the rotating turbine wheel is converted into electrical energy in the electrical generator which transported through the electrical circuit

4) Electrical → Thermal

The electrical energy from the electric current is then converted into thermal energy as the current passes through the resistors in the heating filament

5) Thermal → Thermal

The heated filament transfers thermal energy to the the water in the beaker by conduction which raises the temperature of the water such that as the water acquires more thermal energy it turns into steam

Therefore, we have the flow of energy from the falling water to steam as follows;

1) Mechanical 2) Mechanical 3) Electrical 4) Thermal 5) Thermal

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ECONOMICS !!!!

gulaghasi [49]

Answer:

Explanation:

Absolute advantage is the capability of a commercial entity to produce goods using fewer resources compared to rivals. Using the same inputs, an entity with an absolute advantage produces a larger output compared to competitors. It means the firm has a lower marginal cost of production. Therefore, its products will have the lowest prices in the market.

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3 years ago

Name two methods scientist use to obtain empirical evidence

alisha [4.7K]

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4 years ago

There are many well-documented cases of people surviving falls from heights greater than 20.0 m. In one such case, a 55.0 kg wom

bixtya [17]

1a) -192.7g

1b) 0.0126 s

2) 1309 kg m/s

3) $1.04\cdot 10^5 N$

Explanation:

1a)

First of all, we have to find the velocity of the womena just before hitting the ground.

Since the total mechanical energy is conserved during the fall, the initial gravitational potential energy of the woman when she is at the top is entirely converted into kinetic energy.

So we can write:

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

where

m = 55.0 kg is the mass of the woman

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 29.0 m is the initial height of the woman

v is her final speed

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(29.0)}=23.8 m/s$

Then, when the woman hits the soil, she is decelerated until a final velocity

$v'=0$

So we can find the deceleration using the suvat equation:

$v'^2-v^2=2as$

where

s = 15.0 cm = 0.15 m is the displacement during the deceleration

Solving for a,

$a=\frac{v'^2-v^2}{2s}=\frac{0-23.8^2}{2(0.15)}=-1888.3 m/s^2$

In terms of g,

$a=\frac{-1888.3}{9.8}=-192.7g$

1b)

Here we want to find the time it takes for the woman to stop.

Since her motion is a uniformly accelerated motion, we can do it by using the following suvat equation:

$v'=v+at$

where here we have:

v' = 0 is the final velocity of the woman

v = 23.8 m/s is her initial velocity before the impact

$a=-1888.3 m/s^2$ is the acceleration of the woman

t is the time of the impact

Solving for t, we find:

$t=\frac{v'-v}{a}=\frac{0-23.8}{-1888.3}=0.0126 s$

So, the woman took 0.0126 s to stop.

The impulse exerted on an object is equal to the change in momentum experienced by the object.

Therefore, it is given by:

$I=\Delta p =m(v'-v)$

where

$\Delta p$ is the change in momentum

m is the mass of the object

v is the initial velocity

v' is the final velocity

Here we have:

m = 55.0 kg is the mass of the woman

v = 23.8 m/s is her initial velocity before the impact

v' = 0 is her final velocity

So, the impulse is:

$I=(55.0)(0-23.8)=-1309 kg m/s$

where the negative sign indicates the direction opposite to the motion; so the magnitude is 1309 kg m/s.

The impulse exerted on an object is related to the force applied on the object by the equation

$I=F\Delta t$

where

I is the impulse

F is the average force on the object

$\Delta t$ is the time of the collision

Here we have:

$I=1309 kg m/s$ is the magnitude of the impulse

$\Delta t = 0.0126 s$ is the duration of the collision

Solving for F, we find the magnitude of the average force:

$F=\frac{I}{\Delta t}=\frac{1309}{0.0126}=1.04\cdot 10^5 N$

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4 years ago

How do the prefixes micro,<br> nano and pico relate to each<br> other?

In-s [12.5K]

Answer:

because they are same and their properties

8 0

3 years ago

A free electron is moving with a velocity of 3 × 105 m/s. It strikes a stationary electron in a head-on elastic collision. The m

Gelneren [198K]

Answer:

After the colision, the stationary electron's momentum is given as:

P = 2.7328 x 10^(-25) kg m/s

The direction of momentum is the same as the direction of velocity of the electron.

Explanation:

In an Isolated system, when an object moving at some velocity v collides head on with a stationary object of equal mass. There velocities are exchanged.

This means that the first electron will become stationary and the electron which was stationary initially will start moving at a velocity of 3*10^(5)m/s in the same direction as the first electron.

Post collision momentum of the stationary electron:

V = 3 x 10^5 m/s

m = 9.1093 x 10^(-31) kg

Momentum = P = mV = 9.1093 x 10^(-31) x 3 x 10^5

P = 2.7328 x 10^(-25) kg m/s

The direction of momentum is the same as the velocity of the electron.

7 0

4 years ago

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