<span> Lava </span>erupts<span> on the </span>sea floor<span> also the type of rock that cools quickly is Magma and are finer grained and glassy.
Hope this helps</span>
- rocket science
- automotive research
- space research
Answer:
option D
Explanation:
Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.
Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.
Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.
Hence, the correct answer is option D
Answer:
DS = 13865.7[J/K]
Explanation:
We can calculate the energy of the rock, like the potential energy relative to the lake level. Which can be calculated by means of the following expression of the potential energy:
![E_{p}=m*g*h\\\\where:\\m = mass = 2000[kg]\\h = elevation = 200 [m]\\g = gravity = 9.81[m/s^2]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5C%5C%5Cwhere%3A%5C%5Cm%20%3D%20mass%20%3D%202000%5Bkg%5D%5C%5Ch%20%3D%20elevation%20%3D%20200%20%5Bm%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5Bm%2Fs%5E2%5D)
Therefore:
![E_{p}=2000*9.81*200\\E_{p}=3924000 [J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%3D2000%2A9.81%2A200%5C%5CE_%7Bp%7D%3D3924000%20%5BJ%5D%5C%5C)
This energy is transformed into thermal energy.
we shall remember that isothermal heat transfer processes are internally reversible, so the entropy change of a system during one of these processes can be determined, by the following expression.
![DS=\frac{Q}{T}\\ where:\\DS = entropy change [J/K]\\Q = Heat transfer [J]\\T = temperature [K]](https://tex.z-dn.net/?f=DS%3D%5Cfrac%7BQ%7D%7BT%7D%5C%5C%20where%3A%5C%5CDS%20%3D%20entropy%20change%20%5BJ%2FK%5D%5C%5CQ%20%3D%20Heat%20transfer%20%5BJ%5D%5C%5CT%20%3D%20temperature%20%5BK%5D)
T = 5 + 278 = 283[K]
DS = 3924000 / 283
DS = 13865.7[J/K]
Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,


v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.