<h3><u>Answer;</u></h3>
Continental accretion
<h3><u>Explanation;</u></h3>
- The collision and joining of crustal fragments to a continent is called the continental<em><u> </u></em><em><u>accretion</u></em><em><u>.</u></em>
- <em><u>Accretion is the process through which materials such as sediment, volcanic arcs, or seamounts or blocks, etc. are added to the tectonic plate or land mass.</u></em> Many astronomical bodies, such as stars, galaxies or planets are formed by this process.
- Where the process of accretion has occurred the tectonic plates may have dense basaltic rocks which are indicative of oceanic lithosphere.
Answer:
4334.4 J
Explanation:
Work done equals to kinetic energy change
KE=½mv²
Change in KE is given by
∆KE=½m(v²-u²)
Where m is mass of water-skier, KE is kinetic energy, ∆KE is the change in kinetic energy, v is final velocity and u is initial velocity.
Substituting 72 kg for m, 12.1 m/s for v and 5.10 m/s for u then
∆KE=½*72(12.1²-5.10²)=4334.4J
Therefore, the work done by the net external force acting on the skier is equal to 4334.4 J
Answer:
a. 4 m/s b. 0.2 V
Explanation:
a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.
The hall voltage V = vBd where v = flow-rate, B = magnetic field strength = 0.500 T and d = diameter of pipe = 3.00 cm = 0.03 m
Since V = vBd
v = V/Bd given that V = 60.0 mV = 0.060 V, substituting the values of the other variables, we have
v = 0.060 V/(0.500 T × 0.03 m)
v = 0.060 V/(0.015 Tm)
v = 4 m/s
b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?
Since the hall voltage, V = vBd and v = flow-rate = 4 m/s, B = magnetic field strength = 0.500 T and d' = diameter of pipe = 10.0 cm = 0.10 m
Substituting the variables into the equation, we have
V = vBd
V = 4 m/s × 0.500 T × 0.10 m
V = 0.2 V
Honed I don’t know where the question is
