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Gelneren [198K]
3 years ago
13

A laser beam, shining from the earth's surface, is directed at the moon whose distance from the earth on this day is 370,000 km.

If the beam diverges at an angle of only 1.65 x 10-5 rad, what diameter circle will it make on the moon
Physics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

d = 6105 m

Explanation:

For this exercise we must use trigonometry to find the diameter of the has. The most common definition in optics of the angle of divergence is measured between the two ends of the beam, therefore to use this angle in trigonometry where the angle is measured with respect to the normal we must take half of this angle

          θ = 1.65 10⁻⁵ / 2 = 0.825 10⁻⁵ rad

let's use the tangent

          tan θ = y / L

          y = L tan θ

          y = 370000 103 tan (0.825 10⁻⁵)

let's be careful since the angles are in radians

          y = 3025.5 m

This is the distance from the normal that corresponds to the radius of the circle, the diameter is twice the radius

          d = 2 y

          d = 2 3025.5

          d = 6105 m

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A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
3 years ago
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