It will increase..........................
Explanation:
We have,
Semimajor axis is 
It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :
G is universal gravitational constant
M is solar mass
Plugging all the values,
Since,
So, the orbital period of a dwarf planet is 138.52 years.
Answer:
c. 0.02 C and 4 J
Explanation:
Applying,
Q = CV................ Equation 1
Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.
From the question,
Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V
Substitute these values into equation 1
Q = (50×10⁻⁶)(400)
Q = 0.02 C.
Also Applying
E = CV²/2............. Equation 2
Where E = Energy stored.
Therefore,
E = (50×10⁻⁶ )(400²)/2
E = 4 J
Hence the right option is c. 0.02 C and 4 J
Answer:
Explanation:
Volume of the rectangular block= V = l×b×h
=5cm×5cm×8cm= 200cm³
Density of block= d= 7.5 g/cm³
Mass of the block= m= d× V
= 1500 grams=1.5 kg
Answer:
The temperature of this newly discovered planet violates the third law of thermodynamic, there is a mistake in this value.
Explanation:
The third law of the Thermodynamic says:
<u> At zero kelvin all molecular movement stops, which means that the entropy will be zero at this temperature.</u>
So we can say there is no thermodynamic system that has temperature values less than 0 K.
The conclusion of the report will be.
The temperature of this new planet violates the third law of thermodynamic, there is a mistake in this value.
I hope it helps you!
