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3 years ago

The diagram below shows a periodic wave. Which two points on the wave are 180 degrees out of phase?

Physics

1 answer:

Ksju [112]3 years ago

3 0

Answer:

A and C are 180 deg out of phase (opposite points on a 360 deg wave)

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In the process of changing a flat tire, a motorist uses ahydraulic jack. She begins by applying a force of 45 N to the inputpist

Verdich [7]

Answer:

3100.05 N

Explanation:

$F_1$ = Force on piston = 45 N

$r_1$ = Radius of piston

$r_2$ = Radius of plunger

$\dfrac{r_2}{r_1}=8.3$

From Pascal's law we have

$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}\\\Rightarrow F_2=\dfrac{F_1A_2}{A_1}\\\Rightarrow F_2=\dfrac{F_1\pi r_2^2}{\pi r_1^2}\\\Rightarrow F_2=\dfrac{F_1r_2^2}{r_1^2}\\\Rightarrow F_2=F_1\dfrac{r_2^2}{r_1^2}\\\Rightarrow F_2=45\times 8.3^2\\\Rightarrow F_2=3100.05\ N$

The force is 3100.05 N

3 0

4 years ago

A pressure cylinder has a diameter of 150-mm and has a 6-mm wall thickness. What pressure can this vessel carry if the maximum s

myrzilka [38]

Answer:

p = 8N/mm2

Explanation:

given data ;

diameter of cylinder = 150 mm

thickness of cylinder = 6 mm

maximum shear stress = 25 MPa

we know that

hoop stress is given as = $\frac{pd}{2t}$

axial stress is given as = $\frac{pd}{4t}$

maximum shear stress = (hoop stress - axial stress)/2

putting both stress value to get required pressure

$25 = \frac{ \frac{pd}{2t} -\frac{pd}{4t}}{2}$

$25 = \frac{pd}{8t}$

t = 6 mm

d = 150 mm

therefore we have pressure

p = 8N/mm2

7 0

3 years ago

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:

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

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