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Zanzabum
2 years ago
9

Find an expression for the square of the orbital period.

Physics
1 answer:
jenyasd209 [6]2 years ago
8 0

Answer:

T²= 4π²R³/GM

Explanation:

First we know that

Fg= Fc

Because centripetal force must equal gravitational force

So

GMm/R² = Mv²/R

But velocity is 2πR/T

So by substitution we have

GMm/R²= M (2πR/T)/T

We have

T²= 4π²R³/GM as period

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A car’s 30.0-kg front tire is suspended by a spring with spring constant k=1.00x10^5 N/m. At what speed is the car moving if was
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This is the method by which a wave travels through a medium or a vacuum.
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A thin spherical shell with radius R1 = 2.00 cm is concentric with a larger thin spherical shell with radius R2 = 6.00 cm. Both
Dafna11 [192]

Answer:

a. i. 1350 V ii 0 V iii -450 V b. 6.75 kV. The inner shell is at a higher potential.

Explanation:

The formula for electric potential is given by V = Σkq/r, where k = 9 × 10⁹ Nm²/C², q = charge and r = distance.

q₁ = charge on smaller shell = +6.00 nC = +6.00 × 10⁻⁹ C, r₁ = radius of smaller shell = 2.00 cm = 2.00 × 10⁻² m.

q₂ = charge on larger shell = -9.00 nC = -9.00 × 10⁻⁹ C, r₂ = radius of larger shell = 6.00 cm = 6.00 × 10⁻² m.

a. At r = 0, inside both spheres V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 1350 V

ii. At r = 4.00 cm, the point outside of smaller shell but inside larger shell. r₁ = 4.00 cm = 4.00 × 10⁻² m and r₂ = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁/r₁ + q₂/r₂) = 9 × 10⁹ [+6.00 × 10⁻⁹/4.00 × 10⁻² + (-9.00 × 10⁻⁹/6.00 × 10⁻²)] = 0 V.

iii. At r = 6.00 cm, the point outside both shells. r₁ = r₂ = r = 6.00 cm = 6.00 × 10⁻². So, V = kq₁/r₁ + kq₂/r₂. = k(q₁ + q₂)/r = 9 × 10⁹ [+6.00 × 10⁻⁹+ (-9.00 × 10⁻⁹)]/6.00 × 10⁻² = -450 V.

b. The potential of the surface of the smaller shell is V₁ = 9 × 10⁹ [+6.00 × 10⁻⁹/2.00 × 10⁻²] = 2700 V = 2.7 kV.

The potential of the surface of the larger shell is V₂ = 9 × 10⁹ [-9.00 × 10⁻⁹/2.00 × 10⁻²] = -4050 V = -4.050 kV. The potential difference V₁ - V₂ = 2700 - (-4050) V = 6750 V = 6.75 kV. Since the potential difference is positive, V₁ is higher. So, the inner shell is at a higher potential.

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