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olchik [2.2K]
3 years ago
13

Saturn has an orbital period of 29.46 years. In two or more complete sentences, explain how to calculate the average distance fr

om Saturn to the sun and then calculate it.
Physics
1 answer:
bixtya [17]3 years ago
8 0
For astronomical objects, the time period can be calculated using:
T² = (4π²a³)/GM
where T is time in Earth years, a is distance in Astronomical units, M is solar mass (1 for the sun)
Thus,
T² = a³
a = ∛(29.46²)
a = 0.67 AU
1 AU = 1.496 × 10⁸ Km
0.67 * 1.496 × 10⁸ Km
= 1.43 × 10⁹ Km
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a man weighs 600 n while on the surface of earth. if he is transported to the planet mythos which has the same mass as earth but
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24N

Explanation:

Calculation for what his weight would be

Let his WEIGHT on the surface of earth be 600 and Let the RADIUS be 25 which is (5^²) because we were told that the radius is FIVE TIMES larger than earths

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3 years ago
A circular coil with 600 turns has a radius of 15 cm. The coil is rotating about an axis perpendicular to a magnetic field of 0.
Reil [10]

Answer:

The angular frequency is the coil rotating is 1.9 rad/s.

Explanation:

Given that,

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Radius = 15 cm

Magnetic field = 0.020 T

Maximum induced emf = 1.6 V

We need to calculate the angular frequency is the coil rotating

Using formula emf

\epsilon =NBA\omega\sin\omega t

When  induced emf is maximum so \sin\omega t will be 1.

\omega=\dfrac{\epsilon}{NBA}

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

\omega=\dfrac{1.6}{600\times\pi\times(15\times10^{-2})^2\times0.020}

\omega=1.9\ rad/s

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6 0
3 years ago
A long coaxial cable consists of an inner cylindrical conductor with radius a and an outer coaxial cylinder with inner radius b
Natasha_Volkova [10]

Answer:

Part a)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

Explanation:

Part a)

By Guass law we know that

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part b)

Outside the outer cylinder we will again use Guass law

\int E. dA = \frac{q}{\epsilon_0}

E. 2\pi rL = \frac{\lambda L}{\epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

Part d)

As we know that due to induction of charge there will be same charge appear on the inner and outer surface of the cylinder but the sign of the charge must be different

On the inner side of the cylinder there will be negative charge induce on the inner surface and on the outer surface of the cylinder there will be same magnitude charge with positive sign.

4 0
3 years ago
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