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inna [77]
3 years ago
13

Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 1515 grams and velocity of 77 centimeters pe

r second has a kinetic energy of 147147 ergs. Find the kinetic energy for a mass of 1010 grams and velocity of 99 centimeters per second?
Physics
1 answer:
Alexus [3.1K]3 years ago
4 0

Answer:

The second kinetic energy is 162 J.

Explanation:

Given that,

Mass, m_1=15\ g

Velocity, v_1=7\ cm/s

Kinetic energy, K_1=147\ ergs

Mass, m_2=10\ g

Velocity, v_2=9\ cm/s

We need to find kinetic energy K_2. Kinetic energy is given by :

K=\dfrac{1}{2}mv^2

So,

\dfrac{K_1}{K_2}=\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}\\\\K_2=\dfrac{K_1}{\dfrac{m_1}{m_2}\times \dfrac{v_1^2}{v_2^2}}\\\\K_2=\dfrac{147}{\dfrac{15}{10}\times \dfrac{7^2}{9^2}}\\\\K_2=162\ J

So, the second kinetic energy is 162 J.

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An atom of carbon has a radius of 67.0 pm and the average orbital speed of the electrons in it is about 1.3 x 10⁶ m/s.
adoni [48]

Answer :

The least possible uncertainty in an electron's velocity is, 4.32\times 10^{5}m/s

The percentage of the average speed is, 33 %

Explanation :

According to the Heisenberg's uncertainty principle,

\Delta x\times \Delta p=\frac{h}{4\pi} ...........(1)

where,

\Delta x = uncertainty in position

\Delta p = uncertainty in momentum

h = Planck's constant

And as we know that the momentum is the product of mass and velocity of an object.

p=m\times v

or,

\Delta p=m\times \Delta v      .......(2)

Equating 1 and 2, we get:

\Delta x\times m\times \Delta v=\frac{h}{4\pi}

\Delta v=\frac{h}{4\pi \Delta x\times m}

Given:

m = mass of electron = 9.11\times 10^{-31}kg

h = Planck's constant = 6.626\times 10^{-34}Js

radius of atom = 67.0pm=67.0\times 10^{-12}m     (1pm=10^{-12}m)

\Delta x = diameter of atom = 2\times 67.0\times 10^{-12}m=134.0\times 10^{-12}m

Now put all the given values in the above formula, we get:

\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (134.0\times 10^{-12}m)\times (9.11\times 10^{-31}kg)}

\Delta v=4.32\times 10^{5}m/s

The minimum uncertainty in an electron's velocity is, 4.32\times 10^{5}m/s

Now we have to calculate the percentage of the average speed.

Percentage of average speed = \frac{\text{Uncertainty in speed}}{\text{Average speed}}\times 100

Uncertainty in speed = 4.32\times 10^{5}m/s

Average speed = 1.3\times 10^{6}m/s

Percentage of average speed = \frac{4.32\times 10^{5}m/s}{1.3\times 10^{6}m/s}\times 100

Percentage of average speed = 33.2 % ≈ 33 %

Thus, the percentage of the average speed is, 33 %

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3 years ago
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