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Lana71 [14]
3 years ago
15

APEX What is one advantage of using primary sources when doing research on an

Physics
1 answer:
MissTica3 years ago
4 0

Answer:

C

Explanation:

They are first hand sources so they are more reliable and detailed...

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Can science answer every question
Alla [95]
No because your opinion and beliefs answers many questions
5 0
2 years ago
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What does popular sovereignty
Lana71 [14]

I'm a little confused by your question.

If you mean what is popular sovereignty, it is the belief that the governments is made and sustained by the consent of the people that it governs.

6 0
3 years ago
PROVE THAT G = GM/R² WHERE THE SYMBOLS HAS THEIR USUAL MEANINGS<br>​
butalik [34]

Answer:

g=GM/R^2

Universal Gravutation Constant:

f=GM×m/R^2

Force can be also expressed as

f=m×g

so,

mg=GMxm/R^2

The m gets cancelled so

g=GM/R^2

8 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
DESPERATE WILL GIVE BRAINLSIT AND THANKS
Eva8 [605]

Hello!

My best guess would be hydrogen and oxygen.

hopefully this helps!

8 0
3 years ago
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