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Oksi-84 [34.3K]
3 years ago
13

_ Ba+_O2 =_BaO how do I balance this?

Physics
1 answer:
astraxan [27]3 years ago
5 0

Answer:

02-ba=_4

Explanation:

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Who gathered the data that showed planets traveling in elliptical paths around the sun? who discovered elliptical orbits? who ex
zlopas [31]

Answer:

1. Who gathered the data that showed planets traveling in elliptical paths around the sun?

2. Who discovered elliptical orbits?

3. Who explained them?

Explanation:

1. Tycho Brahe was the one that gathered data.


2. Johannes Kepler discovered it using Tycho Brace’s Data.


3. I believe it was Johannes Kepler because he created 3 laws of it.


I apologize if it is incorrect-

I hope it helps! Have a great day!

Anygays-

3 0
2 years ago
A ball is launched at 40m/s. find its speed at the half way up point.
Andrew [12]
Ek - kinetic energy
v2 - unknown speed
v1 - 40 m/s (initial speed)
Ek=1/2 mv^2
Ek in half way up is 1/2 Ek (with another V)
So, Ek in the beginning is Ek1= 1/2 mv1^2
and in half way Ek2=1/2 mv2^2
Ek1=2*Ek2
1/2 mv1^2 = 2* 1/2 mv2^2
1/2 v1^2 = v2^2
1/2 40^2 = v2^2
800 = v2^2
v2 = sqrt (800) = 28,3 m/s
6 0
3 years ago
Because angular momentum is conserved, an ice-skater who throws her arms out will
ki77a [65]
This is an example of conservation of angular momentum.

The ice skater will rotate more slowly because her arms would be considered the outside torque that is impacting her constant angular momentum.

So, an ice-skater who throws her arms out will rotate more slowly.
8 0
3 years ago
A particle moving along the x-axis has a position given by x = (24t – 2.0t 3 ) m, where t is measured in s. What is the magnitud
Vitek1552 [10]
<h2>The magnitude 24 (\dfrac{m}{s^2} ) of the acceleration of the particle when the particle is not moving.</h2>

Explanation:

Given,

A particle moving along the x-axis has a position given by

x=(24t-2.0t^3) m      ........ (1)

To find, the magnitude (\dfrac{m}{s^2} ) of the acceleration of the particle when the particle is not moving = ?

Differentiating equation (1) w.r.t, 't', we get

\dfrac{dx}{dt} =\dfrac{d((24t-2.0t^3))}{dt}

⇒ \dfrac{dx}{dt} =24(1)-3(2.0)t^{2} =24-6t^{2}     ....... (2)

⇒ 24-6t^{2} = 0

⇒ t^{2}=2^{2}

⇒ t = 2 s

Again, differentiating equation (2) w.r.t, 't', we get

\dfrac{d^2x}{dt^2} =-12t

Put t = 2, we get

\dfrac{d^2x}{dt^2} =-12(2)=24

Thus, the magnitude 24 (\dfrac{m}{s^2} ) of the acceleration of the particle when the particle is not moving.

3 0
3 years ago
Pros and Cons of the Mercator projection as a scientific model.
Andrew [12]

Answer:

ok i don't know the answer but I send you tomorrow

6 0
3 years ago
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