Answer:
5.A mid-ocean ridge or mid-oceanic ridge is an underwater mountain range, formed by plate tectonics. This uplifting of the ocean floor occurs when convection currents rise in the mantle beneath the oceanic crust and create magma where two tectonic plates meet at a divergent boundary.
6.The Nazca plate is an oceanic plate, while the South American plate is continental. The fast moving Nazca plate is moving east towards the South American plate at a downward angle and converging. This process is called subduction, resulting in frequent earthquakes & production of the Andes Mountains.
7.The Nazca plate forms the southeastern part of the Pacific plate. The Nazca and the Pacific plate share both divergent and transform type of plate boundary. The Pacific and the Nazca plate are separating at an increasing rate of about 122-142mm/year.
8.Convection currents in the mantle and in the ocean are similar because they both are responsible for the shaping the Earth's surface. Two forces are behind the movement of Earth's huge land masses. Due to combined action of convection currents and gravity, Earth's plates are in constant motion.
Explanation:
Answer:
A = 2.36m/s
B = 3.71m/s²
C = 29.61m/s2
Explanation:
First, we convert the diameter of the ride from ft to m
10ft = 3m
Speed of the rider is the
v = circumference of the circle divided by time of rotation
v = [2π(D/2)]/T
v = [2π(3/2)]/4
v = 3π/4
v = 2.36m/s
Radial acceleration can also be found as a = v²/r
Where v = speed of the rider
r = radius of the ride
a = 2.36²/1.5
a = 3.71m/s²
If the time of revolution is halved, then radial acceleration is
A = 4π²R/T²
A = (4 * π² * 3)/2²
A = 118.44/4
A = 29.61m/s²
It is itself. This question does not make sense.
Answer:
I = 18 x 10⁻⁹ A = 18 nA
Explanation:
The current is defined as the flow of charge per unit time. Therefore,
I = q/t
where,
I = Average Current passing through nerve cell
q = Total flow of charges through nerve cell
t = time period of flow of charges
Here, in our case:
I = ?
q = (9 pC)(1 x 10⁻¹² C/1 pC) = 9 x 10⁻¹² C
t = (0.5 ms)(1 x 10⁻³ s/1 ms) = 5 x 10⁻⁴ s
Therefore,
I = (9 x 10⁻¹² C)/(5 x 10⁻⁴ s)
<u>I = 18 x 10⁻⁹ A = 18 nA</u>
