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faust18 [17]
4 years ago
13

What happens to potential energy that is not completely converted to kinetic

Physics
1 answer:
Arada [10]4 years ago
7 0

Answer:

There is a possibility that it will be converted to another form, but based on the context the answer is A. It is stored as potential energy.

Explanation:

This is because potential energy means that the energy has the <u>potential</u> to release energy to do work, but it has not yet. So if it has not been transferred to anything, and the object has already had the ability to have potential energy, then it will stay as potential until released.

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A 112 g hockey puck glides across a frictionless ice surface with no horizontal forces acting on it. If the puck's velocity is 2
Burka [1]
22.5 m/s
This is because the puck will not slow down, and maintain a constant velocity.
7 0
4 years ago
1. In what year did India (our second most populous country) begin to produce more energy than Spain?
choli [55]

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

<h2>8) 55 GW </h2>

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

<h2>9) 36 GW </h2>

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

4 0
3 years ago
For how long should a force of 100N act on a body of mass 20kg at rest so that it acquires a velocity of 100m/s?​
julia-pushkina [17]

Answer:

20 seconds

Explanation:

7 0
3 years ago
Read 2 more answers
A star emits electromagnetic radiation which closely resembles the spectrum of a blackbody. The three star system named Albireo
aksik [14]

Answer:

A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm , this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

we have a fluctuation of the intensity emitted by the stars.   Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

Explanation:

The radiation of a black body is characterized by its temperature, with Wien's law of displacement we can find the maximum wavelength emitted by each star.

                  λ T = 2,898 10⁻³

therefore the emission the star of A = 13000K has a maximum at lam = 1,9984 10⁻⁷ m = 199.84 nm

The emission of the premiere is in the ultraviolet light range, as this star is visually separated from the other two by its constant emission spectrum and is not affected by the other two.

The burst with A = 4300K ​​has a bad emission maximum = 6.7395 10⁻⁷ m = 673.95 nm, which corresponds to an emission in the visible in the orange range, giving a blackbody spectrum of this range, but since the emission is formed by two stars, we see that when the two are placed one in front of the other the intensity of the emission must increase significantly and when they are placed next to each other the intensity reaches its minimum, consequently we have a fluctuation of the intensity emitted by the stars.

Consequently by this fluctuation the amateur astronomer can conclude that this system is made up of two stars.

7 0
3 years ago
Which one of the following explains how light energy helps us see all kinds of objects around us?
vampirchik [111]

Answer:

light reflects off the objects and enter our eyes

6 0
3 years ago
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