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3 years ago

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An 8.00 kg point mass and a 15.0 kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point be

tween the two masses 20.0 cm from the 8.00 kg mass along the line connecting the two fixed masses.
Find the magnitude and direction of the acceleration of the particle. Answer from book: 2.2 x 10^-9 m/s^2 - toward the 8.00 kg mass....

1 answer:

Verdich [7]3 years ago

3 0

Answer: $a=2.23\times 10^{-9} m/s^2$

Explanation:

Given

Mass of first Point mass $m_1=8 kg$

Mass of second Point $m_2=15 kg$

distance between them $d=50 cm$

third point mass $m_3=m$

Distance between $m\ and\ m_1 is\ 20 cm$

Distance between $m\ and\ m_2 is 30 cm$

Force Due to $m_1\ and\ m F_1=\frac{Gmm_1}{d_1^2}$

$F_1=\frac{8Gm}{(0.2)^2}$

$F_1=200 mG$

$F_2=m\frac{Gmm_2}{d_2^2}$

$F_1=m\frac{15Gm}{(0.3)^2}$

$F_2=166.67 mG$

Net Force

$F_{net}=F_1-F_2$

$=200 mG-166.67 mG$

$=33.33 mG$

$F_{net}=222.33\times 10^{-11} N$

$F_{net}=2.23m\times 10^{-9} N$

acceleration $a=\frac{2.23m\times 10^{-9}}{m}$

$a=2.23\times 10^{-9} m/s^2$

towards 8 kg mass

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3 years ago

Chapter 14, Problem 042 A flotation device is in the shape of a right cylinder, with a height of 0.588 m and a face area of 4.19

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Answer:

The workdone is $W = 9.28 * 10^{3} J$

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From the question we are told that

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Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

The weight of the cylinder = Upward thrust

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$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

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3 years ago

Two identical satellites orbit the earth in stable orbits. Onesatellite orbits with a speed vat a distance rfrom the center of t

Lelu [443]

Answer:

c) At a distance greater than r

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then orbital speed is ;

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If speed of first satellite = V₁

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If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.

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3 years ago

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 %

3 0

3 years ago