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faltersainse [42]
3 years ago
7

True or False:

Physics
1 answer:
bulgar [2K]3 years ago
8 0

Answer:

False

True

True

True

True

False

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to determine the mass of the central object, we must apply newton's version of kepler's third law, which requires knowing the or
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The approximate orbital period of this star is 13 years.

<h3>What is Kepler's third law?</h3>

The square of a planet's period of revolution around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis, states Kepler's law of periods.

T² ∝ a³

The time it takes for one rotation to complete depends on how closely the planet orbits the sun. With the use of the equations for Newton's theories of motion and gravitation, Kepler's third law assumes a more comprehensive shape:

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2. What are the three phases of water?
Lorico [155]

Answer:

solid, liquid, gas

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3 0
2 years ago
An aluminum wire is held between two clamps under zero tension at room temperature. Reducing the temperature, which results in a
Blababa [14]

Answer:

\frac{\Delta L}{L} =5.37\times 10^{-4}

Explanation:

Given:

  • cross sectional area of the wire, A=5.75\times 10^{-6}\ m^2
  • density of aluminium wire, \rho=2.7\times 10^3\ kg.m^{-3}
  • young's modulus of the material, E=7\times 10^{10}\ N.m^{-2}
  • wave speed, v=118\ m.s^{-1}

<u>We have mathematical expression for strain as:</u>

\frac{\Delta L}{L} =\frac{\sigma}{E} ...............................(1)

and since, \sigma =\frac{T}{A}

where, T = tension force in the wire

equation (1) becomes:

\frac{\Delta L}{L} =\frac{T}{A.E} ............................(2)

<u>Also velocity ofwave in tensed wire:</u>

v=\sqrt{\frac{T}{\mu} } ...................................(3)

where: \mu= linear mass density of the wire

\therefore \mu=\rho \times A

Now, equation (3) becomes

v=\sqrt{\frac{T}{\rho \times A} }

T=v^2.\rho \times A ............................(4)

Using eq. (2) & (4) for tension T

v^2.\rho \times A=A.E\times \frac{\Delta L}{L}

\frac{\Delta L}{L} =\frac{v^2.\rho}{E}

putting the respective values

\frac{\Delta L}{L} =\frac{118^2\times 2.7\times 10^3}{7\times 10^{10}}

\frac{\Delta L}{L} =5.37\times 10^{-4}

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2 years ago
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