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Maksim231197 [3]
3 years ago
7

Gravitational Force: Two small objects, with masses m and M, are originally a distance r apart, and the magnitude of the gravita

tional force on each one is F. The masses are changed to 2m and 2M, and the distance is changed to 4r. What is the magnitude of the new gravitational force?
Physics
1 answer:
Alona [7]3 years ago
4 0

Law of universal gravitation:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m = masses of the objects, r = distance between the objects

F is proportional to both M and m:

F ∝ M, F ∝ m

F is proportional to the inverse square of r:

F ∝ 1/r²

Calculate the scaling factor of F due to the change in M:

k₁ = 2M/M = 2

Calculate the scaling factor of F due to the change in m:

k₂ = 2m/m = 2

Calculate the scaling factor of F due to the change in r:

k₃ = 1/(4r/r)² = 1/16

Multiply the original force F by the scaling factors to obtain the new force:

Fk₁k₂k₃

= F(2)(2)(1/16)

= F/4

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A generator with �# ' = 300 V and Zg = 50 Ω is connected to a load ZL = 75 Ω through a 50-Ω lossless line of length l = 0.15λ. (
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a. Zin = 41.25 - j 16.35 Ω

b. V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

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Explanation:

a.

Zin = Zo * [ ZL + j Zo Tan (βl) ] / [ Zo + j ZL Tan (βl) ]  

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Zin = 50 * [ 75 + j 50 Tan (54) ] / [ 50 + j 75 Tan (54) ]

Zin = 41.25 - j 16.35 Ω

b.

I₁ = Vg / Zg + Zin ⇒ I₁ = 300 / 41.25 - j 16.35 = 3.24 e ¹⁰ ¹⁶

V₁ = I₁ * Zin = 3.24 e ¹⁰ ¹⁶ * ( 41.25 - j 16.35)

V₁ = 143. 6 e⁻ ¹¹ ⁴⁶

c.

Pin = ¹/₂ * Re * [V₁ * I₁]

Pin = ¹/₂ * 143.6 ⁻¹¹ ⁴⁶ * 3.24 e ⁻ ¹⁰ ¹⁶ = 143.6 * 3.24 / 2 * cos (21.62)

Pin = 216 w

d.

The power PL and Pin are the same as the line is lossless input to the line ends up in the load so

PL = Pin

PL = 216 w

e.

Pg Generator

Pg = ¹/₂ * Re * [ V₁ * I₁ ] = 486 * cos (10.16)

Pg = 478.4 w

Pzg dissipated

Pzg = ¹/₂ * I² * Zg = ¹/₂ * 3.24² * 50

Pzg = 262.4 w

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