Describe the location of point a. explain what you used as a reference point for you your location

Physics

1 answer:

Gwar [14]3 years ago

3 0

I used rise and run to see where point a lies .
That's kind of an answer you could you right?

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When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.

3 0

3 years ago

A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease

skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

$P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm$