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kari74 [83]
4 years ago
12

A coal power plant station generates electricity at night when it is not needed. Some of this energy is stored by pumping water

up to a mountain lake. When there is high demand for electricity, the water is allowed to flow back through a turbine to generate electricity.
1- Write down the forms of initial and final energy transfer.

2- On one occasion, 2000 kg of water is pumped up through a vertical height of 500 m. Calculate the gravitational potential energy. (g= 10 m/s^2)
Physics
1 answer:
Dmitry_Shevchenko [17]4 years ago
7 0

Answer:

1.) Check the explanation for the answer

2.) 10 MJ

Explanation:

Given that a coal power plant station generates electricity at night when it is not needed. Some of this energy is stored by pumping water up to a mountain lake. When there is high demand for electricity, the water is allowed to flow back through a turbine to generate electricity.1- Write down the forms of initial and final energy transfer.

Solution.

The initial conversion will be :

Chemical energy is converted to mechanical energy.

The final conversion will be :

Mechanical energy is converted to electrical energy. While the electrical energy is converted to light energy.

2- On one occasion, 2000 kg of water is pumped up through a vertical height of 500 m. Calculate the gravitational potential energy. (g= 10 m/s^2)

The gravitational potential energy can be calculated by using the formula

Energy = mgh.

Where

M = mass

H = height

g = acceleration due to gravity

Substitute all the parameters into the formula

Potential Energy = 2000 × 500 × 10

Potential energy = 10000000 Joule

Potential energy = 10 MJ

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Write an expression to evaluate the relative error in g (given that g= 2h/t^2) in terms of h,Δh,t,Δt
EleoNora [17]

An expression which can be used to evaluate the relative error in g (in terms of h, Δh, t, and Δt) is: \delta = \frac{t^2(h\;+\Delta \;h) }{h(t\;+\;\Delta t)^2} - 1

<h3>What is relative error?</h3>

Relative error can be defined as a measure of the ratio of an absolute (real) value of a measurement to an expected (theoretical) value. Also, it's independent of the magnitude of its values.

<h3>How to evaluate the relative error in g?</h3>

In order to write this expression, we would divide the absolute (real) value by the expected (theoretical) value as follows:

\delta = \frac{g(h\;+\;\Delta h,t \;+ \;\Delta t) - \;g(h,t)}{g(h,t)} \\\\\delta = \frac{g(h\;+\;\Delta h,t \;+ \;\Delta t) }{g(h,t)} - 1\\\\\delta = \frac{t^2(h\;+\Delta \;h) }{h(t\;+\;\Delta t)^2} - 1

<u>Note:</u> g = 2h/t²

Read more on relative error here: brainly.com/question/13370015

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6 0
2 years ago
A youngster throws a rock from a bridge into the river 50 m below. The rock has a speed of 15 m/s when it leaves the youngster’s
butalik [34]
Assuming that the stone is thrown vertically... let's say it's a 1 kg stone.It doesn't matter if it's thrown upwards or downwards as (assuming no air friction) it will pass the original throwing point with the same downwards velocity as it had upwards, 3 seconds previously. So it starts with 1/2 m v^2 = 0.5 * 1 * 15^2 = 112.5 J of keThen k.e. gained = gpe lostk.e. gained = m g h = 1 * 10 * 50 = 500 J of Ke gainedso the final (total) ke is 612.5 J which = 1/2 m v^2 = 0.5 v^2 here
so 0.5 v^2 = 612.5so     v^2    = 1225so v = 35 m/s
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3 years ago
100.0 J of work accelerates a 0.500 kg hockey puck across an ice rink (a frictionless
Olegator [25]

Answer:

d. 100.0 J

Explanation:

To solve this problem we must use the theorem of work and energy conservation. This tells us that the mechanical energy in the final state is equal to the mechanical energy in the initial state plus the work done on a body. In this way we come to the following equation:

E₁ + W₁₋₂ = E₂

where:

E₁ = mechanical energy at state 1. [J] (units of Joules)

E₂ = mechanical energy at state 2. [J]

W₁₋₂ = work done from 1 to 2 [J]

We have to remember that mechanical energy is defined as the sum of potential energy plus kinetic energy.

The energy in the initial state is zero, since there is no movement of the hockey puck before imparting force. E₁ = 0.

The Work on the hockey puck is equal to:

W₁₋₂ = 100 [J]

100 = E₂

Since the ice rink  is horizontal there is no potential energy, there is only kinetic energy

Ek = 100 [J]

It can be said that the work applied on the hockey puck turns into kinetic energy

6 0
3 years ago
An object moves in a straight path. Its position x as a function of time t is presented by the equation x(t) = at – bt2+c, where
ankoles [38]

Answer:

The answer is below

Explanation:

Given that:

x(t) = at – bt2+c

a) x(t) = at – bt2+c

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x(t) = 1.4t - 0.06t² + 50

At t = 5, x(5) = 1.4(5) - 0.06(5)² + 50 = 55.5 m

At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

v=\frac{x(5)-x(0)}{5-0}\\ \\v=\frac{55.5-50}{5-0}=1.1\\ \\v=1.1\ m/s

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At t = 0, x(0) = 1.4(0) - 0.06(0)² + 50 = 50 m

The average velocity (v) is given as:

v=\frac{x(10)-x(0)}{10-0}\\ \\v=\frac{58-50}{10-0}=0.8\\ \\v=0.8\ m/s

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At t = 10, x(10) = 1.4(10) - 0.06(10)² + 50 = 58 m

The average velocity (v) is given as:

v=\frac{x(15)-x(10)}{15-10}\\ \\v=\frac{57.5-58}{15-10}=0.1\\ \\v=0.1\ m/s

6 0
3 years ago
A railroad car of mass m is moving with speed u when it collides with and connects to a second railroad car of mass 3m, initiall
Black_prince [1.1K]
Take a look at the picture. The speed of the connected cars is 1/4 of the single car’s speed, and the kinetic energy is 1/4 of the single car’s kinetic energy

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