B is the correct answer hope that helped
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:
Answer:
The velocity of wind with respect to cyclist is 
.
Explanation:
speed of cyclist = 12 km/h east
speed of wind = 5 km/h south west
Write the speeds in the vector form
The velocity of wind with respect to cyclist is
The axial field is the integration of the field from each element of charge around the ring. Because of symmetry, the field is only in the direction of the axis. The field from an element ds in the ring is
<span>dE = (qs*ds)cos(T)/(4*pi*e0)*(x^2 + R^2) </span>
<span>where x is the distance along the axis from the plane of the ring, R is the radius of the ring, qs is the linear charge density, T is the angle of the field from the x-axis. </span>
<span>However, cos(T) = x/sqrt(x^2 + R^2) </span>
<span>so the equation becomes </span>
<span>dE = (qs*ds)*[x/sqrt(x^2 + R^2)]/(4*pi*e0)*(x^2 + R^2) </span>
<span>dE =[qs*ds/(4*pi*e0)]*x/(x^2 + R^2)^1.5 </span>
<span>Integrating around the ring you get </span>
<span>E = (2*pi*R/4*pi*e0)*x/(x^2 + R^2)^1.5 </span>
<span>E = (R/2*e0)*x*(x^2 + R^2)^-1.5 </span>
<span>we differentiate wrt x, the term R/2*e0 is a constant K, and the derivative is </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 +x*[(-1.5)*(x^2 + R^2)^-2.5]*2x} </span>
<span>dE/dx = K*{(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5} </span>
<span>to find the maxima set this = 0, giving </span>
<span>(x^2 + R^2)^-1.5 - 3*x^2*(x^2 + R^2)^-2.5 = 0 </span>
<span>mult both side by (x^2 + R^2)^2.5 to get </span>
<span>(x^2 + R^2) - 3*x^2 = 0 </span>
<span>-2*x^2 + R^2 = 0 </span>
<span>-2*x^2 = -R^2 </span>
<span>x = (+/-)R/sqrt(2) </span>
Answer: No. Circuit should be closed.
Explanation:
A changing magnetic field induces emf in a circuit which is placed in it. This induced emf causes current to flow through it. This is given by Faraday's law which states that induced emf results due to rate of change of magnetic flux.
Current flows only through a closed circuit. Thus, when a wire is moved up and down in a magnetic field, it will not induce emf because the wire does not form a loop. It is an open circuit.
