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jek_recluse [69]
3 years ago
15

In 1610, galileo used his telescope to discover four prominent moons around jupiter. their mean orbital radii a and periods t ar

e as follows: (a) io has a mean orbital radius of 4.22 x 108 m and a period of 1.77 days. find the mass of jupiter from this information. (b) europa has a mean orbital radius of 6.71 x 108 m and a period of 3.55 days. find the mass of jupiter from this information. (c) ganymede has a mean orbital radius of 10.7 x 108 m and a period of 7.16 days. find the mass of jupiter from this information. (d) callisto has a mean orbital radius of 18.8 x 108 m and a period of 16.7 days. find the mass of jupiter from this information
Physics
1 answer:
katrin2010 [14]3 years ago
6 0

Time period of any moon of Jupiter is given by

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

from above formula we can say that mass of Jupiter is given by

M = \frac{4 \pi^2 r^3}{GT^2}

now for part a)

r = 4.22 * 10^8 m

T = 1.77 day = 152928 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (4.22 * 10^8)^3}{(6.67 * 10^{-11})(152928)^2}

M = 1.9* 10^{27} kg

Part B)

r = 6.71 * 10^8 m

T = 3.55 day = 306720 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (6.71 * 10^8)^3}{(6.67 * 10^{-11})(306720)^2}

M = 1.9* 10^{27} kg

Part c)

r = 10.7 * 10^8 m

T = 7.16 day = 618624 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (10.7 * 10^8)^3}{(6.67 * 10^{-11})(618624)^2}

M = 1.89* 10^{27} kg

PART D)

r = 18.8 * 10^8 m

T = 16.7 day = 1442880 seconds

now by above formula

M = \frac{4 \pi^2 r^3}{GT^2}

M = \frac{4 \pi^2 (18.8 * 10^8)^3}{(6.67 * 10^{-11})(1442880)^2}

M = 1.889* 10^{27} kg

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lorasvet [3.4K]

Answer:

(a). The amount of current that flows through this portion of the membrane is 2.70\times10^{-12}\ A

(b). The factor of the current is increase by factor of 2.

(1) is correct option

Explanation:

Given that,

Thickness = 7.50 nm

Area A=(1.3\times1.3\times10^{-6})^2

Potential difference = 92.2 mV

Resistivity of the material \rho=1.30\times10^{7}\ \ohm

We need to calculate the resistance

Using formula of resistivity

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.30\times10^{7}\times7.50\times10^{-9}}{(1.3\times1.3\times10^{-6})^2}

R=3.413\times10^{10}\ \Omega

(a). We need to calculate the amount of current that flows through this portion of the membrane

Using Ohm's law

V=IR

I=\dfrac{V}{R}

Put the value into the formula

I=\dfrac{92.2\times10^{-3}}{3.413\times10^{10}}

I=2.70\times10^{-12}\ A

The amount of current that flows through this portion of the membrane is 2.70\times10^{-12}\ A

(b). If the side dimensions of the membrane portion is halved

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R'=\dfrac{\rho \dfrac{l}{2}}{A}

Put the value into the formula

R'=\dfrac{1.30\times10^{7}\times\dfrac{7.50\times10^{-9}}{2}}{(1.3\times1.3\times10^{-6})^2}

R'=1.706\times10^{10}\ \Omega

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Using Ohm's law

V=I'R'

I'=\dfrac{V}{R'}

Put the value into the formula

I'=\dfrac{92.2\times10^{-3}}{1.706\times10^{10}}

I'=5.404\times10^{-12}\ A

We need to calculate the factor

\dfrac{I'}{I}=\dfrac{5.404\times10^{-12}}{2.70\times10^{-12}}

I'=2I

The factor of the current is increase by factor of 2.

Hence,(a). The amount of current that flows through this portion of the membrane is 2.70\times10^{-12}\ A

(b). The factor of the current is increase by factor of 2.

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A bug is on the rim of a disk of diameter 9 in that moves from rest to an angular speed of 76 rev/min in 6.5 s. what is the tang
ehidna [41]
Let's convert the final angular speed of the disk into rad/s:
\omega_f = 76  \frac{rev}{min} \cdot  \frac{2 \pi rad/rev}{60 s/min}=7.96 rad/s
while the initial angular speed is zero.

So, the angular acceleration of the disk is
\alpha =  \frac{\omega_f - \omega_i }{t} = \frac{7.96 rad/s}{6.5 s}=1.22 rad/s^2

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And so, the tangential acceleration is the angular acceleration times the radius:
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Water absorbs and releases large amount of energy before changing temperature, a characteristic known as
Anton [14]
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If the slope of a line on a speed-time graph is 0 and it is not on the x-axis, what is the object doing?
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With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t
VladimirAG [237]

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu  = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v =  \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N

Therefore, the tension of the rope is 34.95 N

4 0
3 years ago
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