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matrenka [14]
3 years ago
12

Convert one mean solar day into second​

Physics
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

86400 seconds

Explanation:

To convert a mean solar day to second :

Number of hours in a solar day = 24 hours

Number of minutes per hour = 60 minutes

Number of seconds per minute = 60 minutes

Hence, the number of seconds in a solar day is :

(Number of hours in a solar day * number of minutes in an hour * number of seconds in a minute)

(24 * 60 * 60) seconds = 86400 seconds

Hence, 1 mean solar day = 86400 seconds

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3 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

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6 0
3 years ago
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Which factors affect heat transfer between a warm and a cool substance?
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Answer: B

Explanation:

It's not the time it took to heat the substance, so that rules out A and C.

This means that we only have to choose between

B. the area of contact

D. the area of the substances

(since everything else in each of those answers are the same)

Area of contact matters more (e.g. an object with greater surface area is exposed to the air more, will lose/gain heat quicker than an object with less surface area).

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2 years ago
Problem 8 I estimate that the Gauss gun (a solenoid) is wound with 500 turns over a distance of 15cm with an average radius of 1
stellarik [79]

Answer:

Energy stored, U = 66.6 J

Explanation:

It is given that,

Number of turns in the solenoid, n = 500

Radius of solenoid, r = 1.5 cm = 0.015 m

Distance, d = 15 cm = 0.15 m

Let U is the energy stored in the solenoid. Its formula is given by :

U=\dfrac{1}{2}LI^2

L is the self inductance of the solenoid

L=\mu_o N^2A d

N is the no of turns per unit length

L=\mu_o (n/d)^2A d

L=\dfrac{4\pi\cdot10^{-7}\cdot500^{2}\cdot\pi\cdot\left(0.015\right)^{2}}{0.15}

L = 0.00148 Henry

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U = 66.6 J

Out of given options, the correct option for the energy stored in the solenoid is 70 J. So, the correct option is (a) "70 J".  

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3 years ago
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