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Tamiku [17]
3 years ago
14

The diameter of the Moon is 3480 km. What is the surface area of the Moon?

Physics
2 answers:
valentinak56 [21]3 years ago
8 0
You will need to convert the diameter to radius. 3840km diameter will be 1740km radius. 

<span>4 x 3.1416 x 1740² = ~38,046,032km² surface area.</span>
Lelu [443]3 years ago
5 0

Answer:

Surface area of the moon is S=3.8\times 10^{13}\ m^2

Explanation:

It is given that,

Diameter of the moon, d = 3480 km

Radius of the moon, r=1740\ km=1740\times 10^3\ m

Moon is in the form of sphere. The surface area of the moon is given by :

S=4\pi r^2

S=4\pi \times (1740\times 10^3)^2

S=3.8\times 10^{13}\ m^2

So, the surface area of the moon is 3.8\times 10^{13}\ m^2. Hence, this is the required solution.

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3 years ago
A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

        x = 0.135 cm

Given parameters

  • The mass m = 220 g = 0.220 kg
  • The spring cosntnate3 k = 7.0 N / m
  • Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

  • The position where the kinetic and potential energy are equal

 

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

          x = A cos wt + fi

          w² = \frac{k}{m}

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

       v = \frac{dx}{dt}

let's evaluate

       v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

       0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

        x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

        K = U

        ½ m v² = m g x

       

we substitute

        ½ A² w² sin² wt = g A cos wt

        sin² wt = \frac{2g}{A}  cos wt

let's calculate

      w = \sqrt{\frac{7}{0.220} }  

      w = 5.64 rad / s

      sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

      sin² 5.64t = 376.92 cos 5.64 t

      1 - cos² 5.64t = 376.92 cos 5.64t

      cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

       x = cos 5.64t

      x²- 376.92 x - 1 = 0

      x = 0.026

      cos 5.64t = 0.026

   

Let's find the displacement for this time

       x = 5.2 10-2 0.026

       x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

        x = 0.135 cm

Learn more here: brainly.com/question/15707891

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