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sergey [27]
3 years ago
6

most responsible for the formation of a star is most responsible for the formation of a star is blank Force

Physics
1 answer:
Bad White [126]3 years ago
5 0
That would be gravitational force. Check out creation.com. Search up "star-formation" in their search bar. Hope this helped.
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Using a power supply, I place +1.00µC on one plate of a capacitor. A capacitor is a device for storing charge. I, then, connect
Paraphin [41]

Answer:

N=62421972534 electrons

Explanation:

The charge transferred charge Q= −10nC = -10*10-9 C

Charge of a electron : Q_{e}= -1,602*10^{-19}C

Number of electrons transferred:  N=\frac{Q}{Q_{e} }=(-10*10^{-9} )/(-1,602*10^{-19})=62421972534 electrons

5 0
3 years ago
Scientists need to know how to make measurements
Verizon [17]

Part of the scientific process involves sharing your results with other scientists. To do this, we all need to use the same measurement system, which you'll learn about in this lesson.

Imagine you're trying to find out how much an elephant weighs. You're pretty sure it weighs a lot, but you don't know the exact number. So you ask your teacher, and she tells you an elephant weighs the same as three hippos.

Well that's nice to know, but how much does a hippopotamus weigh? Again, you ask your teacher, and she tells you a hippopotamus weighs the same as five alligators. That's a cool fact to know, but you still don't understand how much an elephant weighs because comparing elephants to alligators can be confusing.

plz mark me as brainliest :)

4 0
3 years ago
A car initially at rest undergoes uniform acceleration for 6.32 seconds and covers a distance of 120 meters. What is the approxi
Verizon [17]
Using kinematic equation s=ut + 1/2 at^2(u = initial velocity=0, s=120m, t= 6.32s), 120 = 0(t) + 1/2 a(6.32)^2. a = 120x2/(6.32)^2 = 6m/s^2.  
3 0
3 years ago
Read 2 more answers
Which of the following is an example of refraction?
Sladkaya [172]
I believe it will be C.
7 0
2 years ago
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Find an expression for the electric field E⃗ at the center of the semicircle. Hint: A small piece of arc length Δs spans a small
Hatshy [7]

Answer:

Electric Field a the centreE=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

Explanation:

<u>Given:</u>

Total charge on the semicircle =Q

Radius of the semicircle=R

Let consider a elemental charge on the semicircle at an angle \theta\\ with the horizontal. Since the The charge distribution is symmetrical about y axis so the there will symmetrical charge on other side. The horizontal component will cancel out each other and vertical component will get added so let dE be the electric Field due to elemental charge

Let \lambda be the charge per unit length such that\lambda=\dfrac{Q}{\pi R}

k=\dfrac{1}{4\pi \epsilon_0}

Total Electric Field at the centre

=2dE\sin\theta\\=2\dfrac{k\lambda }{R}\int \sin \theta d\theta\\

integrating 0 to \dfrac{\pi}{2}

E=\dfrac{2k\lambda}{R}(-\vec j)

E=\dfrac{Q}{2\pi^2\epsilon_0 R^2}(-\vec j)

So the Electric field at the centre is calculated.

7 0
3 years ago
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