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3 years ago

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A of mass 5.0 kilograms moving at 20. meters per second collides with ball B of unknown mass moving at 10. meters per second in

the same direction. After the collision, ball A moves at 10. meters per second and ball B at 15 meters per second, both still in the same direction. What is the mass of ball B

1 answer:

barxatty [35]3 years ago

7 0

Answer:

Mass of B is 10 kg

Explanation:

M1v1a + m2v1b = m1v2a + m2v2b

Let m2 = x

5*20 + x10 = 5*10 + x15

100 + x10 = 50 + x15

50 = x5

X= 10

Mass of B is 10 kg

M1 = mass of A

V1a = initial velocity of A

V2a = final velocity of A

M2 = mass of B

V1b = initial velocity of B

V2b = final velocity of B

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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

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2 years ago

By what factor is the self-inductance of an air solenoid changed if its length and number of coil turns are both tripled

fredd [130]

Answer:

The new self inductance is 3 times of the initial self inductance.

Explanation:

The self inductance of a solenoid is given by :

$L=\dfrac{\mu_oN^2 A}{L}$

Where

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A is area of cross section

l is length of solenoid

If length and number of coil turns are both tripled,

l' = 3l and N' = 3N

New self inductance is given by :

$L'=\dfrac{\mu_oN'^2 A}{L'}\\\\=\dfrac{\mu_o(3N)^2 A}{3L}\\\\=3\dfrac{\mu_oN^2 A}{L}\\\\=3L$

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You are asked to explain what happens to individual atoms when a chemical change takes place in lab you are doing in class. What

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2 years ago

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3 years ago

PLEASE HELP ME WITH THIS ONE QUESTION

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Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

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[A] = Our incognite

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k = ln 2 / Half-life

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<h3>c) 2.02 x 10^16 nuclei</h3>

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2 years ago