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3 years ago

1. In what year did India (our second most populous country) begin to produce more energy than Spain?

Physics

1 answer:

choli [55]3 years ago

4 0

Answers:

<h2>1) Sometime in 2014 </h2>

As we can see in the graph in 2014 there is an intersection between the lines that represent the wind energy production of Spain and India, then the production of India is increased, while that of Spain remains constant.

<h2>2) Germany </h2>

As we can see in the graph in 2000 Germany was the number 1 producer of gigawatts of wind energy. In addition it is observed a rapid growth and increase in production compared to the other countries.

<h2>3) In 2006, America was producing over 10 GW and began a steep climb in wind energy production.</h2>

In 1997, America began with a constant and very small energy production, which was growing gradually until 2006, when its rapid growth in production began to exceed the other countries shown in the graph.

<h2>4) Years</h2>

When a function is plotted, generally the independent variable, whose value is previously set, is represented on the X axis.

In this sense, the variable found on the X axis of this graph is Years.

<h2>5) Wind Energy Capacity </h2>

When a function is plotted, generally the dependent variable (which is generated from the independent variable) is represented on the Y axis.

In this sense, the variable found on the Y axis of this graph is Wind Energy Capacity.

<h2>6) Gigawatts (GW) </h2>

We already know the dependent variable in this <u>Wind Energy Capacity vs Years</u> graph is Wind Energy Capacity, and its Unit is Gigawatts (GW).

Where $1GW=10^{6} W$

<h2>7) Both onshore and offshore wind sources </h2>

When we talk about scientific experimentation and representing data, the Control Variable is the one that remains constant (it does not change during the investigation).

According to the explanation above, the option that fits with this characteristic is: Both onshore and offshore wind sources

If we want to make a linear interpolation to estimate how much wind energy capacity was there in the United States in the middle of 2011, firstly we need to <u>find two points where the value we want to find is in the middle.</u>

These points (X,Y) are:

P1: (2011,48) and P2: (2012,60)

Where X1:2011, Y1:48, X2:2012, Y2:60

Now we find the <u>slope</u> $m$ of the line with the following equation:

$m=\frac{Y2 - Y1}{X2 - X1}$

$m=\frac{60 - 48}{2012 - 2011}=12$

Then, with this value of the slope we can write the <u>equation of the line</u> and find the <u>intersection point</u> $b$ with one of the given points (we will use P1):

$Y=mX+b$

$48=12(2011)+b$

$b=-24084$

With this value and the slope, we can find Y for X=2011.5 (we need to know the wind capacity in the middle of 2011):

$Y=(12)(2011.5)-24084$

$Y=54$ This is the e<u>stimated result</u> 54 GW, and the option that is near this value is 55 GW

In this case we will apply a similar method as the previous answer, but estimating a prediction:

Let's find two points near the value we want to find. These points (X,Y) are:

P1: (2014,22) and P2: (2016,29)

Where X1:2014, Y1:22, X2:2016, Y2:29

Now we find the <u>slope</u> $m$ of the line with the following equation:

$m=\frac{Y2 - Y1}{X2 - X1}$

$m=\frac{29 - 22}{2016 - 2014}=\frac{7}{2}$

Then, with this value of the slope we can write the <u>equation of the line</u> and find the <u>intersection point</u> $b$ with one of the given points (we will use P2):

$Y=mX+b$

$29=\frac{7}{2}(2016)+b$

$b=-7027$

With this value and the slope, we can find Y for X=2020 (we need to estimate the expected wind capacity in 2020):

$Y=\frac{7}{2}(2020)-7027$

$Y=43$ This is the e<u>stimated result</u> 43 GW (note linear extrapolation is not as accurate as other methods), and the option that is near this value is 36 GW

<h2>10) Additional 7 GW/Year</h2>

In this case we will use the equation of the slope:

$m=\frac{Y2 - Y1}{X2 - X1}$

With the following points:

P1: (2009,35) and P2: (2016,82)

$m=\frac{82 - 35}{2016 - 2009}$

$m=6.7 \approx 7$

Therefore, the correct option is Additional 35 GW/Year

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A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri

exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

$U_i +K_i = U_f + K_f$

where

$U_i$ is the initial potential energy, at the top

$K_i$ is the initial kinetic energy, at the top

$U_f$ is the final potential energy, at halfway

$K_f$ is the final kinetic energy, at halfway

The equation can be rewritten as

$mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2$

where:

m = 2.7 kg is the mass of the block

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

$h_i = 0.54$ is the initial height

u = 0 is the initial speed

$h_f = 0.25 m$ is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

$v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s$

The total mechanical energy of the block can be calculated from the initial conditions, and it is

$E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J$

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

$E=E_e = \frac{1}{2}kx^2$

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

$x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m$

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

$E=E_e = 14.3 J$