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2 years ago

11

The distance and displacement of a object in motion can be the same (true or false)

1 answer:

Rasek [7]2 years ago

4 0

Aswer:

False, the values of the distance traveled and the displacement only coincide when the trayectorie is a straight line. Otherwise, the distance will always be greater than the offset.

Although these terms are used synonymously in other cases, they are totally different. Since the distance that a mobile travels is the equivalent of the length of its trajectory. Whereas, the displacement will be a vector magnitude.

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pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

<u>For m = 7 [g] and V = 2.3 [cm³]</u>

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

<u>For m = 10 [g] and V = 7 [cm³]</u>

<u /> $Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$ <u />

<u>For m = 15 [g] and V = 5 [cm³]</u>

<u /> $Ro=15/5\\Ro=3[g/cm^{3} ]\\$ <u />

<u>For m = 21 [g] and V = 8 [cm³]</u>

<u /> $Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$ <u />

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2 years ago

What amount of energy is needed for an electron to jump from n = 1 to n = 4?

liberstina [14]

Answer:

$E=2.04\times 10^{-18}\ J$

Explanation:

We need to find the energy for an electron to jump from n = 1 to n = 4.

The energy in transition from 1 state to another is given by :

$E=\dfrac{-2.18\times 10^{-18}}{n^2}\ J$

The difference in energy for n = 1 to n = 4 is:

$E=-2.18\times 10^{-18}\times (\dfrac{1}{4^2}-1)\\\\E=2.04\times 10^{-18}\ J$

So, the required energy is equal to $2.04\times 10^{-18}\ J$ .

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Answer:

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Explanation:

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Romashka-Z-Leto [24]

Answer:

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Explanation:

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