Answer:The Atwood Machine is a device that demonstrates the basic principles of acceleration and dynamics. You'll mostly see Atwood machines in Physics laboratories and classrooms. It consists of two objects with different masses that hang vertically from a frictionless pulley that has a very small, negligible mass.
Explanation:
Metal ores
Explanation:
in an area where subduction has occurred in times past, metal ores are likely to be found.
Metallic ores find subduction zone regions very favorable to crystallize out of a magma.
- Ores have different modes of formation.
- Typically, they are found in hydrothermal vents and black smokers of igneous intrusives.
- These are igneous terrains where metallic sulfides and other minerals crystallize out of magmatic body.
- Metals in magma usually have large sizes and do not partition easily in the melt.
At a subduction zone, partial melting of the subducting plate forces magma into nearby country rock as an intrusive and to the ocean floor where they form black smokers.
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The period T is time it takes for one complete cycle or from "trough to trough" so the reverse is trough per sec = 1/T = frequency
Answer: Long Answer...
Explanation: Most of the time, chemical energy is released in the form of heat, and this transformation from chemical energy to heat, or thermal energy, is called an exothermic reaction. Next, there are two main types of mechanical energy: kinetic energy and potential energy.
Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) =
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ =
–K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J