Incomplete Question.The Complete question is
The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.
(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?
Answer:
(i) KE= 2.56e29 J
(ii) KE= 2.65e33 J
Explanation:
i) Treating the Earth as a solid sphere, its moment of inertia about its axis is
I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²
I = 9.69e37 kg·m²
About its axis,
ω = 2π rads/day * 1day/24h * 1h/3600s
ω= 7.27e-5 rad/s,
so its rotational kinetic energy
KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²
KE= 2.56e29 J
(ii) About the sun,
I = mR²
I= 5.97e24kg * (1.496e11m)²
I= 1.336e47 kg·m²
and the angular velocity
ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s
ω= 1.99e-7 rad/s
so
KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²
KE= 2.65e33 J
Answer:
W = - 118.24 J (negative sign shows that work is done on piston)
Explanation:
First, we find the change in internal energy of the diatomic gas by using the following formula:
where,
ΔU = Change in internal energy of gas = ?
n = no. of moles of gas = 0.0884 mole
Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)
Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K
ΔT = Rise in Temperature = 18.8 K
Therefore,
Now, we can apply First Law of Thermodynamics as follows:
where,
ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)
W = Work done = ?
Therefore,
<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>
The answer is 0.025J.
W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J
Reduce friction because friction just makes things harder
