The answer you are looking for is B, hope this helps.
Answer:
In general, the annual sea surface temperatures(SSTs) in the Bay of Bengal(BOB) are higher than the Arabian sea(AS). because, there are two main reasons for higher SST in the Bay of Bengal than the Arabian Sea. they are 1. stratification, 2.strong mixing
stratification is nothing but a phenomenon which stratifies(layers) the sea water when different density water(fresh water, rain water) add into the sea water. So the stratification in the bay of Bengal is comparatively high than the Arabian sea due to the high river discharge and precipitation in the BOB than the AS. the mixing process over the Arabian sea is higher than the Bay of Bengal due to the prevailing of strong winds over the AS (because of the presence of the mountains of east Africa) than Bay of Bengal (because of the winds over the BOB are sluggish in nature then the AS). But generally winds over the sea mixes easily the normal sea water than stratified/stabilized sea water column. That's why less mixing will takes place over the surface of BOB than the AS. So due to the presence of less mixing over the surface of the Bay of Bengal than the Arabian sea, the SST values over the Arabian sea are always lower than the Bay of Bengal. that's why the Arabian sea is colder than the Bay of Bengal.
Explanation:
Oil leakage in aquatic areas. (fish)
Burning fuel. (Air/global warming)
Toxic waste burried in dirt (living Creatures)
Lights/Electricity (destroyes Birds migration)
Noise (migration on dolphines and other sea mammals)
Answer:
speed of the charge electric is v = - (Eo q/m) cos t
Explanation:
The electric charge has a very small mass so it follows the oscillations of the electric field. We force ourselves on the load,
F = q Eo sint
a) To find the velocity of the particle, let's use Newton's second law to find the acceleration and of this by integration the velocity
F = ma
q Eo sint = ma
a = Eo q / m sint
a = dv / dt
dv = adt
∫ dv = ∫ a dt
v-vo = I (Eoq / m) sin t dt
v- vo = Eo q / m (-cos t)
We evaluate the integral from the initial point, as the particle starts from rest Vo = 0, for t = 0
v = - (Eo q / m) cos t
b) Kinetic energy
K = ½ m v2
K = ½ m (Eoq / m)²2 (sint)²
K = ¹/₂ Eo² q² / m sin² t
c) The average kinetic energy over a period
K = ½ m v2
<v2> = (Eoq / m) 2 <cos2 t>
The average of cos2 t = ½, substitute and calculate
K = ½ m (Eoq / m)² ½
K = ¼ Eo² q² / m
