A planet or butting a distant star has been observed to have an orbital period of 0.76 earth years at a distance of 1.2 au. What
1 answer:
Answer:
The mass of the star is, M = 5.9567x10³⁰ Kg
Explanation:
Given
The orbital period of the planet, T = 0.76 year
= 2.3967x10⁷ seconds
The distance between planet and sun, R+h = 1.2 a.u
= 1.795 x 10¹¹ meters
The orbital period of the planet is given by the formula
Squaring and solving for M
Substituting the given values in the above equation
M = 5.9567 x 10³⁰ Kg
Hence, the mass of the star the planet is orbiting, M = 5.9567 x 10³⁰ Kg
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Answer:
(a) Charge of 4.25 μF capacitor is 35.46 μC.
(b) Charge of 1.13 μF capacitor is 10.05 μC.
(c) Charge of 2.85 μF capacitor is 25.36 μC.
Explanation:
Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:
C₄ = C₁ + C₂
Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.
C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF
Since, C₄ and C₃ are connected in series, there equivalent capacitance is:
C₅ =
Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.
C₅ =
C₅ = 2.05 μF
The charge on the equivalent capacitance is determine by the relation :
Q = C₅ V
Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.
Q = 2.05 μF x 17.3 = 35.46 μC
Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.
Charge on the capacitor, C₃ = 35.46 μC
Charge on the capacitor, C₄ = 35.46 μC
Voltage on the capacitor C₄ = =
= 8.90 volts
Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.
Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC
Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC
Answer: A. 23.59 minutes.
B. 249.65 minutes
Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.
<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).
<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.
Here, <u>given that</u>:
- mass of ice, m= 0.555 kg
- temperature of ice, T= -16.6°C
- rate of heat transfer,
- specific heat of ice,
- latent heat of fusion of ice,
<u>Asked:</u>
1. Time require for the ice to start melting.
2. Time required to raise the temperature above freezing point.
Sol.: 1.
<u>We have the formula:</u>
Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.
Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.
∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.
= 23.59 minutes.
Sol.: 2.
Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.
<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>
From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.
<u>We have the equation:</u> latent heat,
<u>Now the time required for supply of 185370 J:</u>
t= 226.06 minutes
∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59
= 249.65 minutes