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3 years ago

Make ansentance rkdloebebjekeoejbe

Physics

2 answers:

forsale [732]3 years ago

8 0

Answer:

<em>just did by typing this lol </em>

Arada [10]3 years ago

6 0

Answer:

the man has returned from his trip

You might be interested in

Part 1

wel

Answer:

Part 1: 0.3789

Part 2: 746 J

Part 3: 2.162 kW

Explanation:

Part 1:

Eff= $1-\frac{1223}{1969}$

Eff= 0.378873 ≈ 0/3789

Part 2:

W= 0.3789(1969)

W= 746 J

Part 3:

$Power=\frac{W}{t}$

$Power= \frac{746}{0.345}$

Power= 2162.3188 Watts

2162.3188 W-----> 2.162 kW

4 0

2 years ago

How would the force between two charged particles change if one of the charges were to triple in strength (3x stronger)?

Andreyy89

Answer:

The correct option is (d). "3x stronger".

Explanation:

The force between two charged particle is given by :

$F=k\dfrac{q_1q_2}{r^2}$

If one charge is tripled, $q_1'=3q$

New force will be :

$F'=\dfrac{kq_1'q_2}{r^2}\\\\=3\times F$

It means if one of the charges were to triple in strength, then the force will become 3 times of the initial force.

8 0

3 years ago

How was the Periodic Table of Elements developed and how are the elements arranged on it?

victus00 [196]

Answer:

In 1869 Russian chemist Dimitri Mendeleev started the development of the periodic table, arranging chemical elements by atomic mass. He predicted the discovery of other elements, and left spaces open in his periodic table for them.

Explanation:

4 0

3 years ago

Ms. Mann's experiment illustrates several important facts in science

torisob [31]

Answer:

(B)

Explanation:

I did it in USATestprep.

8 0

3 years ago

A tire is filled with air at 22oC to a gauge pressure of 240 kPa. After driving for some time, if the temperature of air inside

german

Answer:

7.8% of the original volume.

Explanation:

From the given information:

Temperature $T_1$ = 22° C = 273 + 22 = 295° C

Pressure $P_1$ = 240 kPa

Temperature $T_2$ = 45° C

At initial temperature and pressure:

Using the ideal gas equation:

$P_1V_1 =nRT_1$

making V_1 (initial volume) the subject:

$V_1 = \dfrac{nRT_1}{P_1}$

$V_1 = \dfrac{nR*295}{240}$

Provided the pressure maintained its rate at 240 kPa, when the temperature reached 45° C, then:

the final volume $V_2$ can be computed as:

$V_2 = \dfrac{nR*318}{240}$

Now, the change in the volume ΔV = V₂ - V₁

$\Delta V = \dfrac{nR*318}{240}- \dfrac{nR*295}{240}$

$\Delta V = \dfrac{23nR}{240}$

∴

The required fraction of the volume of air to keep up the pressure at (240) kPa can be computed as:

$= \dfrac{\dfrac{23nR}{240}}{ \dfrac{295nR}{240}}$

$= {\dfrac{23nR}{240}} \times { \dfrac{240}{295nR}}$

$= 0.078$

= 7.8% of the original volume.

3 0

3 years ago

Make ansentance rkdloebebjekeoejbe​

Make ansentance rkdloebebjekeoejbe