If a circuit contains a 12v battery and two 3 bulbs in series

1 answer:

6 0

If the resistance of each bulb is 3 ohms, and they are connected across the battery, then current will flow through them.

In series, the two 3-ohm bulbs will look like a single 6-ohm resistance.

Current = (voltage) / (resistance) = (12v) / (6 ohms) = 2 Amperes.

The battery will need to deliver some power to the circuit.

Power = (voltage) x (current) = (12v) x (2 Amperes) = 24 watts .

Each bulb will dissipate 12 watts, in the form of light and heat.

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Consider the following geometric solids.

Bad White [126]

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 $0.08m^{-1}$

The surface area to volume ratio for right circular cylinder is given $2.1m^{-1}$

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.

5 0

3 years ago

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At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$