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:Energy cannot be created or destroyed. It can be stored, or it can be transferred i.e. from a hot object to a cool object by conduction, convection or radiation.
Explanation:
Answer:
13.33m/s
Explanation:
Given data
m1= 2000kg
u1= 20m/s
m2= 1500kg
u2= 0m/s
v1= 10m/s
Required
The speed of the sticks
We know that from the expression for the conservation of momentum
m1u1+m2u2= m1v1+m2v2
2000*20+1500*0=2000*10+1500*v2
40000=20000+1500v2
collect like terms
40000-20000= 1500v2
20000= 1500v2
v2= 20000/1500
v2= 13.33 m/s
Hence the velocity of the sticks is 13.33m/s
Our solar system is the only place in the universe where gravity played a key part in the formation of planets.
This is Flase
Answer:its D
Explanation:
i did the test and got it right