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mart [117]
2 years ago
13

Which of the following correctly describes the result of a voltage difference?

Physics
2 answers:
Maslowich2 years ago
7 0
The answer is d charges flow from high voltage areas to low voltage areas
Goryan [66]2 years ago
5 0

Answer:

D from high to low areas

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Electrical metallic tubing would be used in an electrical installation because it <span>is less expensive.</span>
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How many additional valence electrons does fluorine need to have a full
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Answer:

D= Seven

Explanation:

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For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t
Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at (\frac{35}{2},0)

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

y=2+\frac{1}{x}

y=2+x^{-1}

y'=-x^{-2}

y'=-\frac{1}{x^{2}}

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(1)^{2}}

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-3=-1(x-1})

y-3=-1x+1

y=-x+1+3

y=-x+4

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-x+4=0

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(2)^{2}}

m=y'=-\frac{1}{4}

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-2.5=-\frac{1}{4}(x-2})

y-2.5=-\frac{1}{4}x+\frac{1}{2}

y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}

y=-\frac{1}{4}x+3

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-\frac{1}{4}x+3=0

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(2.5)^{2}}

m=y'=-\frac{4}{25}

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-2.4=-\frac{4}{25}(x-2.5})

y-2.4=-\frac{4}{25}x+\frac{2}{5}

y=-\frac{4}{25}x+\frac{2}{5}+2.4

y=-\frac{4}{25}x+\frac{14}{5}

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-\frac{4}{25}x+\frac{14}{5}=0

and solve for x

x=\frac{35}{20}

so, when fired from (2.5, 2.4) the rocket will hit the target at (\frac{35}{2},0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

y'=-\frac{1}{x^{2}}

y'=-\frac{1}{(4)^{2}}

m=y'=-\frac{1}{16}

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

y-y_{1}=m(x-x_{1})

y-2.25=-\frac{1}{16}(x-4})

y-2.25=-\frac{1}{16}x+\frac{1}{4}

y=-\frac{1}{16}x+\frac{1}{4}+2.25

y=-\frac{1}{16}x+\frac{5}{2}

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

-\frac{1}{16}x+\frac{5}{2}=0

and solve for x

x=40

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0
3 years ago
What two things are necessary for work to be done on an object?
NemiM [27]

Answer: Force and Movement

Explanation:

The first is that the object moves. The second is that a force must act on the object in the direction the object moves.

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All of the noble gases, Group 18, have eight valence electrons in its outer shell (excluding helium which only has two). Which o
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Answer:

Zero or +2

Explanation:

The noble gases already have a avplete outermost shell. They are the least reactive elements of earth?

Their normal oxidation number is zero but some have been shown to be reactive.

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