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Allisa [31]
3 years ago
6

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

Physics
1 answer:
padilas [110]3 years ago
3 0

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

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A horizontal spring with spring constant 85 N/m extends outward from a wall just above floor level. A 5.5 kg box sliding across
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3 years ago
Assume that the body's muscle mechanism can be approximated by a spring with a uniform continuous mass distribution that follows
tatiyna

Based on Hooke's law, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

<h3>What is the spring constant?</h3>

The spring constant or stiffness constant of an elastic spring is constant which describes the extent a bit forceapplied to an elastic spring will extend it.

  • Spring constant, K = force/extension

Assuming, a body's muscle mechanism is a spring obeying Hooke's law, the effective mass of the spring with mass m is 1/3 of the mass of the spring = m/3

The potential energy that can be stored = ke^2 / 2

where K is spring constant and e is the extension produced.

Therefore, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

Learn more about Hooke's law at: brainly.com/question/12253978

5 0
2 years ago
If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi
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Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

F_s = \mu_s*N

Where F_s is the static friction force, \mu_s is the coefficient of static friction and N is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, N = 110\text{ N}, to make the puppy moving we need to use a force of 80 N, therefore, F_s = 80 \text{ N}, so we can solve for the coefficient as shown below:

80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\

The coefficient of static friction between the puppy and the floor is 0.7273.

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3 years ago
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