It is overhead at the equator, it is because the sun ray’s
will be moving vertically as this will be directed at the equator. It is
because if it moves vertically, it will hit or overhead the equator and this
usually happens in spring and fall.
Answer:
4A
Explanation:
According to ohm's law;
E = IRt where;
E is the source voltage = 24volts
I is the total current flowing in the circuit = ?
Rt is the total effective resistance in the circuit.
To find Rt, we will resolve the resistors in parallel first.
Since 6ohms and 12ohms resistors are in parallel, their effective resistance will give;
1/R = 1/6+1/12
1/R= 2+1/12
1/R = 3/12
3R = 12
R = 4ohms.
This resistor will now be in series with the 2.0ohms resistor to finally have;
Rt = 4+2
Rt = 6ohms
From the ohms law formula;
I = E/Rt
I = 24/6
I = 4Amperes
The total current in the circuit is 4A
This same currents will flow in the 2ohms resistor since same current flows in a series connected resistors.
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given 
The surface area to volume ratio for right circular cylinder is given 
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
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