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Finger [1]
2 years ago
11

Nanotechnology is currently used in electronics for

Physics
1 answer:
VMariaS [17]2 years ago
3 0

Answer:

data storage, OLEDS, and wires

Explanation:

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What happens when the fuel in a car engine combustion?​
Trava [24]

In a spark ignition engine, the fuel is mixed with air and then inducted into the cylinder during the intake process. After the piston compresses the fuel-air mixture, the spark ignites it, causing combustion. The expansion of the combustion gases pushes the piston during the power stroke.

Credit: Department of Energy

5 0
2 years ago
A girl and boy pull in opposite directions on a stuffed animal. The girl exerts a force of 3.5 N. The mass of the stuffed animal
Dmitry [639]
The resultant force on the animal = Resultant mass * total acceleration 
F = 0.2 * 2.5 to the right
F = 0.5 to the right.

As, girl exerting a force of 3.5 N & it's not mentioned that she is in right or left, so the force exerting by boy would be either:
3.5-0.5 = 3  OR  3.5+0.5 = 4

If boy exerting a greater force then, answer will be 4 N & if girl exerting a greater force the, answer will be 3 N

Hope this helps!
3 0
3 years ago
Read 2 more answers
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
Physical values in the real world have two components: magnitude and
Nadusha1986 [10]

Answer:

Dimension.

Explanation:

6 0
3 years ago
A speed boat increases its speed uniformly from vi = 20.0 m/s to vf = 30.0 m/s in a distance of 2.00 x 10^2m. (a) Draw a coordin
pychu [463]

a) See graph in attachment

b) The suvat equation to use is v_f^2 - v_i^2 = 2as

c) The acceleration is a=\frac{v_f^2-v_i^2}{2s}

d) The acceleration is 1.25 m/s^2

e) The time needed is 8 s

Explanation:

a)

For this part, find in attachment the diagram representing this situation.

Since we are not given any particular direction for the motion, we choose the x-direction as the direction of motion of the boat.

Then we have the following:

- The initial position of the boat is x_i = 0, the origin

- The  final position of the boat is x_f = 200 m

- The initial velocity of the boat is v_i = 20.0 m/s

- The final velocity of the boat is v_f = 30.0 m/s

Note that the arrow representing the final velocity is longer than that of the initial velocity, since the final velocity is larger.

b)

The motion of the speed boat is a uniformly accelerated motion (motion at constant acceleration), therefore we can use one of the suvat equations. In this particular problem, we know the following quantities:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

Therefore, the equation that best can be use to find the acceleration is

v_f^2 - v_i^2 = 2as

where

a is the acceleration

c)

Now we have to solve the equation

v_f^2 - v_i^2 = 2as

In order to find the acceleration.

This can be done by dividing both terms by 2s: this way, we find

\frac{v_f^2-v_i^2}{2s}=\frac{2as}{2s}

And so the acceleration is

a=\frac{v_f^2-v_i^2}{2s}

d)

Now we can use the equation found in part c) in order to find the acceleration.

We have the following data:

v_i = 20.0 m/s, the initial velocity

v_f = 30.0 m/s, the final velocity

s = x_f - x_i = 200 m, the  displacement of the boat

And substituting into the equation,

a=\frac{30^2-20^2}{2(200)}=1.25 m/s^2

e)

In order to find the time it takes the boat to travel the given distance, we can use the following suvat equation:

v_f = v_i + at

where:

v_i is the initial velocity

v_f is the final velocity

a is the acceleration

t is the time

Here we have:

v_i = 20.0 m/s

v_f = 30.0 m/s

a=1.25 m/s^2

Solving for t, we find:

t=\frac{v_f-v_i}{a}=\frac{30-20}{1.25}=8 s

Learn more about accelerated motion:

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#LearnwithBrainly

8 0
3 years ago
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