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3 years ago

What is the primary function of earth's atmosphere (1 point)?

1 answer:

5 0

<span>Without the atmosphere that protects the Earth, humans, animals and all plant life would not be able to survive. The atmosphere works to hold oxygen in the area around the earth. Humans and most animals are dependent on oxygen. Plants are also dependent on oxygen because they take the leftover carbon dioxide from humans when they inhale the oxygen.</span>

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I need someone smart!

Ivanshal [37]

Answer:

If we are looking for evidence of something that exists outside of our visible Universe and leaves no trace within it, it seems that the idea of a Multiverse is fundamentally untestable. But there are all sorts of things that we cannot observe that we know must be true. Decades before we directly detected gravitational waves, we knew that they must exist, because we observed their effects.

Explanation:

Maybe helps lol

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3 years ago

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Which description accurately describes the tide represented by the image below?

igomit [66]

<span>The gravitational pull of the sun and moon combined
create larger than normal tides.</span>

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3 years ago

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What is the wavelength of a wave that has a speed of 160 m/s and a period of 3.7 ms?

Elza [17]

Answer:

160m/s

Explanation:

The speed of a wave is related to its frequency and wavelength, according to this equation:

v=f ×λ

3 0

2 years ago

Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?

Alik [6]

Answer:

Below

Explanation:

You can use this equation to find the distance :

distance = velocity x time

distance = (26.7)(3.06)

= 81.702 m

Rounding to 3 sig figs

= 81.7 m

Hope this helps

6 0

3 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

4 years ago