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o-na [289]
4 years ago
14

What would happen if the planets would go out of orbit and go on their own (10 Points)

Physics
1 answer:
Ilia_Sergeevich [38]4 years ago
8 0
I would say that the apmosphere would change and we probley would die that's my guess

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Which of the following describes the effect of a convex mirror on light rays?
natulia [17]
B. It reflects light rays outward.

A convex mirror looks like this ⊂.  So when light rays hit the outward curves, it bends the light rays outwards, and backwards,  and it would appear to be coming from its center called the focus.

3 0
3 years ago
HURRYYYY pls help due today n i be giving brainliest!!!<br><br><br> n no mfkn links plsss
nordsb [41]

Answer:

1 Frequency

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3 years ago
In the following figure, electric field at y axis will be maximum at y=?
Strike441 [17]

Because of symmetry electric field component in the x axis cancels out. Now just use electric field formula and slap that sine of theta cause you want the vertical component of electric field and multiply that by two since there’s two charges. I’ve shown my work. Hope it helps✌

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3 years ago
Read 2 more answers
A hill that has a 28.1% grade is one that rises 28.1 m vertically for every 100.0 ml of distance in the horizontal direction. At
Serhud [2]

Answer:

\theta=15.70^\circ

Explanation:

A right triangle is formed, in which the vertical elevation is the opposite cathetus and the horizontal distance is the adjacent cathetus, since we know these two values, we can calculate the angle of inclination using the definition of tangent:

tan\theta=\frac{opp}{adj}\\\theta=arctan(\frac{opp}{adj})\\\theta=arctan(\frac{28.1m}{100m})\\\theta=15.70^\circ

6 0
4 years ago
A regulation basketball has a 32 cm diameter
Rudik [331]

Answer:

1.8 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 4.8 m sin 39.4°)

v = 5.47 m/s

The acceleration down the incline is constant, so given:

Δx = 4.8 m

v₀ = 0 m/s

v = 5.47 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (4.8 m) / (5.47 m/s + 0 m/s)

t = 1.76 s

Rounding to two significant figures, it takes 1.8 seconds.

3 0
4 years ago
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