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3 years ago

8

If you shout into the Grand Canyon,your voice travels at a speed of 340 m/s to the bottom of the canyon and back,and you hear an

echo.How deep is the Grand Canyon at the spot where you hear an echo 5.2 seconds after?

1 answer:

schepotkina [342]3 years ago

7 0

The grand canyon is 1,857 meters deep.

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Dr. Potter provides vaccinations against polio and measles.

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Answer:

hi zzayn i will solve this later

Explanation:

Let's establish that x= number of polio vaccinations and y=number of measles vaccinations.

Since Dr. P gave a total of 60 VACCINATIONS, we know that x+y=60. (This is equation #1)

We also know that Dr. P gave a total of 184 DOSES of the vaccinations and that the polio vaccine has 4 doses, while the measles vaccine has 2 doses. This means that our second equation is 4x+2y=184. (This is equation #2)

You can get either variable (x or y) by itself from equation #1. I always choose x.

x+y=60

-y. -y

x=60-y

Now we can plug our x value into equation #2.

4x+2y=184

4(60-y)+2y=184

240-4y+2y=184

240-2y=184

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y=28 measles vaccines

So we know that Dr. P administered 28 measles vaccines. We need to plug our y value back into equation #1 (or, to make it easier, our switched around equation) to find the number of polio vaccines.

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Answer:

the angle is about 67.79 degrees

Explanation:

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We also know that the maximum height of the projectile is given by the square of its initial vertical component of the velocity (Vyi) divided by 2g, therefore half of such distance is :

$half\,\,max-height = \frac{v_{yi}^2}{4\,g}$

we can use this information to find the y component of the velocity at that height via the formula:

$v_{yf}^2-v_{yi}^2=-2\,g\,\Delta y\\\\v_{yf}^2-v_{yi}^2=-2\,g\,(\frac{v_{yi}^2}{4\,g} )\\v_{yf}^2=v_{yi}^2-\frac{v_{yi}^2}{2} \\v_{yf}^2=\frac{v_{yi}^2}{2}$

Now we use the information that tells us the speed of the projectile at this height to be 1 m/s. That should be the result of the vector addition of the vertical and horizontal components:

$1=\sqrt{v_{yf}^2+0.5^2} \\1=\sqrt{\frac{v_{yi}^2}{2} +0.5^2}\\1^2=\frac{v_{yi}^2}{2} +0.5^2}\\1-0.5^2=\frac{v_{yi}^2}{2} \\2(1-0.5^2)=v_{yi}^2\\1.5 = v_{yi}^2\\v_{yi}=\sqrt{1.5} \\$

Now we can use the arc-tangent to calculate the launching angle, since we know the two initial component of the velocity vector:

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Answer:

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Explanation:

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