Answer:
Winds tend to rotate in a counter clockwise direction in the center of northern and southern hemisphere.
Explanation:
The wind blows clockwise around a high pressure area in the northern hemisphere and the wind blows counter - clockwise around low pressure.
In the northern hemisphere High-pressure systems rotate clockwise direction and in the southern hemisphere low-pressure systems rotate clockwise direction.
If one of two interacting charges is doubled, the force between the charges will double.
Explanation:
The force between two charges is given by Coulomb's law

K=constant= 9 x 10⁹ N m²/C²
q1= charge on first particle
q2= charge on second particle
r= distance between the two charges
Now if the first charge is doubled,
we get 
F'= 2 F
Thus the force gets doubled.
Answer:
(A) It will take 22 sec to come in rest
(b) Work done for coming in rest will be 0.2131 J
Explanation:
We have given the player turntable initially rotating at speed of 
Now speed is reduced by 75 %
So final speed 
Time t = 5.5 sec
From first equation of motion we know that '

(a) Now final velocity 
So time t to come in rest 
(b) The work done in coming rest is given by

Answer:

Explanation:
P = Acoustic power = 63 µW
r = Distance to the sound source = 210 m
Acoustic power

Threshold intensity = 
Ratio

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68
Answer:

Explanation:
Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.
Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t
A(0) = 10 g
Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min
The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min
But the concentration is total amount of salt over 350L constant volume
C = A / 350
Therefore our rate of change for salt A' is
A' = 5 - 5A/350 = 5 - A/70
This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

So 
with A(0) = 10
c + 350 = 10
c = 10 - 350 = -340
