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LuckyWell [14K]
2 years ago
11

A certain satellite travels in an approximately circular orbit of radius 2.0 × 106 m with a period of 7 h 11 min. Calculate the

mass of its planet from this information.
Physics
1 answer:
kap26 [50]2 years ago
5 0

Answer: Mass of the planet, M= 8.53 x 10^8kg

Explanation:

Given Radius = 2.0 x 106m

Period T = 7h 11m

Using the third law of kepler's equation which states that the square of the orbital period of any planet is proportional to the cube of the semi-major axis of its orbit.

This is represented by the equation

T^2 = ( 4π^2/GM) R^3

Where T is the period in seconds

T = (7h x 60m + 11m)(60 sec)

= 25860 sec

G represents the gravitational constant

= 6.6 x 10^-11 N.m^2/kg^2 and M is the mass of the planet

Making M the subject of the formula,

M = (4π^2/G)*R^3/T^2

M = (4π^2/ 6.6 x10^-11)*(2×106m)^3(25860s)^2

Therefore Mass of the planet, M= 8.53 x 10^8kg

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The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. what percentage of the mechanical energy of th
Zanzabum

E = ½KA^2 is the mechanical energy of any oscillator.  It is the sum of elastic potential energy and kinetic energy.  When amplitude A decreases by 3%, then

(E2-E1)/E1 = {½K(A2^2/A1^2) }/ ½K(A1^2)

= {(A2^2 – A1^2) / (A1^2)}

= 97^2 – 100^2/100^2

= 5.91% of the mechanical energy is lost each cycle.

3 0
2 years ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Aleksandr-060686 [28]

Answer:

Explanation:

We have the following relation between power, P and intensity, I

I = P/(4*pi*r^2)

= 10^3/(4*pi*(35000*10^3))

= 6.5*10^-14 W/M^2

We also have the following relationship between electric field and electromagnetic radiation thus

I = (ceE^2)/2

Hence E = \sqrt{2I/ce}

substituting the values of I, c and e, we have

7*10^-6 V/m

3 0
2 years ago
A large piece of jewelry has a mass of 132.6 g. a graduated cylinder initially contains 48.6 ml water. when the jewelry is subme
Lerok [7]

The density of the substance is<u> 10.5 g/cm³.</u>

The jewelry is made out of <u>Silver.</u>

Density ρ is defined as the ratio of mass <em>m</em> of the substance to its volume V<em>. </em> The cylinder contains a volume <em>V₁ of water</em> and when the jewelry is immersed in it, the total volume of water and the jewelry is found to be V₂.

The volume <em>V</em> of the jewelry is given by,

V=V_2 -V_1

Substitute 48.6 ml for <em>V₁ </em>and 61.2 ml for V₂.

V=V_2 -V_1\\ =61.2 ml -48.6 ml\\ =12.6 ml

calculate the density ρ of the jewelry using the expression,

\rho =\frac{m}{V}

Substitute 132.6 g for <em>m</em> and 12.6 ml for <em>V</em>.

\rho =\frac{m}{V}\\ =\frac{132.6 g}{12.6 ml} \\ =10.5 g/ml

Since 1 ml=1 cm^3,

The density of the jewelry is <u> 10.5 g/cm³.</u>

From standard tables, it can be seen that the substance used to make the jewelry is <u>silver</u><em><u>, </u></em>which has a density 10.5 g/cm³.



4 0
2 years ago
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 30 mph. at the same tim
aleksandr82 [10.1K]

In triangle ABC , using Pythagorean theorem

BC = sqrt(AB² + AC²)

r = sqrt(y² + x²)                                eq-1

taking derivative both side relative to "t"

dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)

15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)

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3 0
2 years ago
Read 2 more answers
A uniformly charged spherical droplet of mercury has electric potential Vbig throughout the droplet. The droplet then breaks int
mezya [45]

Answer:

\frac{V_{big}}{V_{small}} = n^{2/3}

Explanation:

Let the total charge on the big drop is given as Q

now if the radius of the drop is R then electric potential of the big drop is given as

V_{big} = \frac{KQ}{R}

Now if it break into n identical drops

then let the charge on each drop is "q" and radius is "r"

by volume conservation

\frac{4}{3}\pi R^3 = n(\frac{4}{3}\pi r^3)

r = \frac{R}{n^{1/3}}

now we have potential of smaller drop given as

V_{small} = \frac{kq}{r}

V_{small} = \frac{K(Q/n)}{\frac{R}{n^{1/3}}}

V_{small} = \frac{1}{n^{2/3}}\frac{KQ}{R}

\frac{V_{big}}{V_{small}} = n^{2/3}

7 0
2 years ago
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