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adell [148]
2 years ago
9

With a bit of algebraic reasoning find your gravitational acceleration toward any planet of mass M a distance d from its center.

Physics
1 answer:
grandymaker [24]2 years ago
5 0

The acceleration due to gravity is given as:

                             g = GM/r²

<h3>Derivation of gravitational acceleration:</h3>

According to Newton's second law of motion,

F = ma

where,

F = force

m = mass

a = acceleration

According to Newton's law of gravity,

F<em>g </em>= GMm/(r + h)²

F<em>g = </em>gravitational force

From Newton's second law of motion,

F<em>g </em>= ma

a = F<em>g</em>/m

We can refer to "a" as "g"

a = g = GMm/(m)(r + h)²

g = GM/(r + h)²

When the object is on or close to the surface, the value of g is constant and height has no considerable impact. Hence, it can be written as,

g = GM/r²

Learn more about gravitational acceleration here:

brainly.com/question/2142879

#SPJ4

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Answer: They have different rigidities.

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An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

                    =          1.224 x 10¹⁴ 

                    =          122,400 GHz          <== my calculation

                                      11.1 MHz           <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation. 
  











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