Auroras are frequently seen : B. After solar flares
The Aurora is created by an ongoing influx of particles into the Earth's existing magnetic field,
This particles originated from the Sun as part of Solar wind
hope this helps
Answer:
Explanation:
Let the velocity be v
Total energy at the bottom
= rotational + linear kinetic energy
= 1/2 Iω² + 1/2 mv² ( I moment of inertia of shell = mr² )
= 1/2 mr²ω² + 1/2 mv² ( v = ω r )
= 1/2 mv² +1/2 mv²
= mv²
mv² = mgh ( conservation of energy )
v² = gh
v = √gh
= √9.8 x 1.8
= 4.2 m /s
There are different options here but all of them work by approximating and assuming.
i) that the boulder is above ground.
ii) that the bottom surface of the boulder is known.
iii) the shape of the boulder is taken into account.
The most accurate way is measuring it by displacement method but the boulder is immovable hence the volume can be calculated by measuring the boulder or a waterproof box to be built around the boulder and calculate the volume occupied by boulder.
All the above methods are estimating methods.
*Another way to find the density is through specific gravity.
S.G = <u>Density</u><u> </u><u>of</u><u> </u><u>object</u>
Density of water
If the material that makes the boulder is known that is if it's stone or a mineral then the specific gravity can be found.
If the boulder is purely rock then S.G lies between 3 - 3.5 and the density of water is known thus the density of the boulder can be found without moving the boulder.
This is what I think after correction and allthe best!
Answer:
Explanation:
The volume of a sphere is:
V = 4/3 * π * a^3
The volume charge density would then be:
p = Q/V
p = 3*Q/(4 * π * a^3)
If the charge density depends on the radius:
p = f(r) = k * r
I integrate the charge density in spherical coordinates. The charge density integrated in the whole volume is equal to total charge.





Since p = k*r
Q = p*π^2*r^3 / 2
Then:
p(r) = 2*Q / (π^2*r^3)
1. The balls move to the opposite direction but the same speed. This represents Newton's third law of motion.
2. The total momentum before and after the collision stays constant or is conserved.
3. If the masses were the same, the velocities of both balls after the collision would exchange.
4 and 5. Use momentum balance to solve for the final velocities.