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Murljashka [212]
3 years ago
10

Newton’s law of universal gravitation a. is equivalent to Kepler’s first law of planetary motion. b. can be used to derive Keple

r’s third law of planetary motion. c. can be used to disprove Kepler’s laws of planetary motion. d. does not apply to Kepler’s laws of planetary motion.
Physics
2 answers:
Finger [1]3 years ago
5 0
Kepler derived his three laws of planetary motion entirely from
observations of the planets and their motions in the sky.

Newton published his law of universal gravitation almost a hundred
years later.  Using some calculus and some analytic geometry, which
any serious sophomore in an engineering college should be able to do,
it can be shown that IF Newton's law of gravitation is correct, then it MUST
lead to Kepler's laws.  Gravity, as Newton described it, must make the planets
in their orbits behave exactly as they do.

This demonstration is a tremendous boost for the work of both Kepler
and Newton.
Nikolay [14]3 years ago
5 0
<h3><u>Answer;</u></h3>

b. can be used to derive Kepler’s third law of planetary motion.

Newton’s law of universal gravitation <u><em>can be used to derive Kepler’s third law of planetary motion</em></u>.

<h3><u>Explanation</u>;</h3>

<u><em>Kepler's</em></u> three laws of planetary motion states that;

  • <em><u>All planets move about the Sun in elliptical orbits, having the Sun as one of the foci. </u></em>
  • <em><u>A radius vector joining any planet to the Sun sweeps out equal areas in equal lengths of time.</u></em>
  • <em><u>The squares of the sidereal periods of the planets are directly proportional to the cubes of their mean distances from the Sun</u></em>

<em><u>Newton's law of universal gravitation</u></em><em><u> states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.</u></em>

<u><em>Newton’s law of universal gravitation can be used to derive Kepler’s third law of planetary motion.</em></u>

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Sliva [168]
Double displacement...I think
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3 years ago
The two measurements necessary for calculating average speed are
alisha [4.7K]

The correct answer is Option (C) distance and time

Explanation:

Average speed of any object is defined as the total distance that object travels over the time it takes to travel that distance. In other words, average speed is the total distance divided by the elapsed time.

Average \thinspace Speed = \frac{Total \thinspace Distance}{Elapsed \thinspace Time}

Therefore, as you can see in the above equation, the two measurements that are essential for the calculation of the average speed are the (total) distance and the (elapsed) time.

Hence, the correct option is C.

5 0
3 years ago
Determine the magnitude of the resultant force and its direction using both the parallelogram and Cartesian vector notation meth
Alika [10]

Answer:

   F = 1494.52 N,   θ = 44º

Explanation:

For the sum of vectors by the parallelogram method, see attached, the vectors are drawn, the parallelogram is completed and a vector is drawn from the origin of the two vectors to the end point of the rectangle, this is the resulting vector.

The attachment shows this roughly.

For the Cartesian coordinate method, each vector is decomposed into its components, they are added algebraically and then the resulting vector is composed in the form of a module and angles

we use trigonometry to decompose the vectors.

The coordinate system can be seen in the attachment

           sin θ = y / R

           cos θ = x / R

            y = R sin θ

            x = R cos θ

Vector 1

module F₁ and angle β₁ = 50

            sin 50 = \frac{F_{1y} }{F_1}

            cos 50 = \frac{F_{1x} }{F_1}

            F_{1y} = F₁ sin 50

            F₁ₓ = F₁ cos 50

            F_{1y} = 600 sin 50 = 459.63 N

            F₁ₓ = 600 cos 50 = 385.67 N

Vector 2

modulus F₂ = 900N, angle β₂ = 40

            F_{2y} = 900 sin 40 = 578.51 N

            F₂ₓ = 900 cos 40 = 689.44 N

we find the resultant of each component

           F_{y} =F_{1y} + F_{2y}

           F_{y}  = 459.63 + 578.51

           F_{y}  = 1038.14 N

 

            Fₓ = F₁ₓ + F₂ₓ

            Fₓ = 385.67 + 689.44

             Fₓ = 1075.11 N

We use the Pythagorean theorem to find the modulus of the resultant

            F = Fₓ² + F_{y}^2

            F = √(1075.11² + 1038.14²)

            F = 1494.52 N

we use trigonometry for the angle

            tan θ = F_y / Fₓ

            θ = tan⁻¹ (F_y / Fₓ)

            θ = tan⁻¹ (1038.14 / 1075.11)

            θ = 44º

8 0
2 years ago
How does height affects the potential energy of a body?<br>​
ivann1987 [24]

Answer:

The higher the height above the more PE (Potential Energy), the body posses as seen in the formula PE = mass*gravity*height. m and g are constant hence only h affect PE

Explanation:

Potential Energy (PE) is the energy the body posses by virtue of its vertical height above the ground because, once the body is released, it will release or create some energy eg hydroelectricity power generation or a car going downhill does not need to be accelerated by the driver since it can roll over on its own due to PE

3 0
3 years ago
What can be the maximum value of the original kinetic energy of disk AA so as not to exceed the maximum allowed value of the the
timurjin [86]

The question is incomplete. The complete question is :

In your job as a mechanical engineer you are designing a flywheel and clutch-plate system. Disk A is made of a lighter material than disk B, and the moment of inertia of disk A about the shaft is one-third that of disk B. The moment of inertia of the shaft is negligible. With the clutch disconnected, A is brought up to an angular speed ?0; B is initially at rest. The accelerating torque is then removed from A, and A is coupled to B. (Ignore bearing friction.) The design specifications allow for a maximum of 2300 J of thermal energy to be developed when the connection is made. What can be the maximum value of the original kinetic energy of disk A so as not to exceed the maximum allowed value of the thermal energy?

Solution :

Let M.I. of disk A = $I_0$

So, M.I. of disk B =  $3I_0$

Angular velocity of A = $\omega_0$

So the kinetic energy of the disk A = $\frac{1}{2}I_0\omega^2$

After coupling, the angular velocity of both the disks will be equal to ω.

Angular momentum will be conserved.

So,

$I_0\omega_0 = I_0 \omega + 3I_0 \omega$

$I_0\omega_0 = 4I_0 \omega$

$\omega = \frac{\omega_0}{4}$

Now,

$K.E. = \frac{1}{2}I_0\omega^2+ \frac{1}{2}3I_0\omega^2$

$K.E. = \frac{1}{2}I_0\frac{\omega_0^2}{16}+ \frac{1}{2}3I_0\frac{\omega_0^2}{16}$

$K.E. = \frac{1}{2}I_0\omega_0^2 \left(\frac{1}{16}+\frac{3}{16}\right)$

$K.E. = \frac{1}{2}I_0\omega_0^2\times \frac{1}{4}$

$\Delta K = \frac{1}{2}I_0\omega_0^2 - \frac{1}{2}I_0\omega_0^2 \times \frac{1}{4} $

$2300=\frac{3}{4}\left(\frac{1}{2}I_0\omega_0^2\right)$

$\frac{1}{2}I_0\omega_0^2=2300 \times \frac{4}{3 } \ J $

Therefore, the maximum initial K.E. = 3066.67 J

3 0
3 years ago
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