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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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(b) 0.012V kWh

(c) 0.108V cents

Explanation:

<u>Given:</u>

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  • t = time interval for which the current flow = 4\ h = 4\times 3600\ s = 14400\ s
  • V = terminal voltage of the battery
  • R = rate of energy = 9 cents/kWh

<u>Assume:</u>

  • Q = charge transported as a result of charging
  • E = energy expended
  • C = cost of charging

Part (a):

We know that the charge flow rate is the electric current flow through a wire.

\therefore i = \dfrac{Q}{t}\\\Rightarrow Q =it\\\Rightarrow Q = 3\times 14400\\\Rightarrow Q = 43200\ C\\\Rightarrow Q = 43.200\ kC\\

Hence, 43.2 kC of charge is transported as a result of charging.

Part (b):

We know the electrical energy dissipated due to current flow across a voltage drop for a time interval is given by:

E = Vit\\\Rightarrow E = V\times 3\times 4\\\Rightarrow E = 12V\ Wh\\\Rightarrow E = 0.012V\ kWh\\

Hence, 0.012V kWh is expended in charging the battery.

Part (c):

We know that the energy cost is equal to the product of energy expended and the rate of energy.

\therefore \textrm{Cost}=\textrm{Energy}\times \textrm{Rate}\\\Rightarrow C = ER\\\Rightarrow C = 0.012V\times 9\\\Rightarrow C =0.108V\ cents

Hence, 0.108V cents is the charging cost of the battery.

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3 years ago
A blacksmith heats a 1.5 kg iron horseshoe to 500 C, containing 20 kg of water at 18 C then plunges it into a bucket What is the
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Answer:

21.85 C

Explanation:

mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C

mass of water = 20 kg, initial temperature of water, T2 = 18 C

let T be the equilibrium temperature.

Specific heat of iron = 449 J/kg C

specific heat of water = 4186 J/kg C

Use the principle of caloriemetry

heat lost by the hot body = heat gained by the cold body

mass of iron x specific heat of iron x decrease in temperature = mass of water  x specific heat of water x increase in temperature

1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)

336750 - 673.5 T = 83720 T - 1506960

1843710 = 84393.5 T

T = 21.85 C

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