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3 years ago

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Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at

8.08 m/s, and Ethan, mass 31.0 kg, is gliding to the left at 10.9 m/s along the same line. When they meet, they grab each other and hang on.(a) What is their velocity immediately thereafter?

1 answer:

Maurinko [17]3 years ago

5 0

Answer:

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob

Explanation:

Given:

mass of Jacob, $m_j=45\ kg$

velocity of Jacob, $v_j=8.08\ m.s^{-1}$

mass of Ethan, $m_e=31\ kg$

velocity of Ethan, $v_e=10.9\ m.s^{-1}$

Now using the conservation of linear momentum for the case:

(When the two masses in motion combine to form one after the collision then they will move together in the direction of the greater momentum.)

$(m_j+m_e).v=m_j.v_j-m_e.v_e$

$(45+31)\times v=45\times 8.08-31\times 10.9$

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob as it was assumed to be positive.

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To find the magnitude, the given values are,

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<h3>What is magnitude?</h3>

The term magnitude can be defined as “ quantity ”.
For instance, the magnitude can be used for describing about the comparison of speeds.

It can also be used to explain the distance travelled by an object or to explain the amount of an object in terms of its magnitude.
Magnitude of the velocity vector is total value which is square root of the value.

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1 year ago

Draw the vector c⃗ =1.5a⃗ −3b⃗

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<span>The magnitude of a is 1.5
The magnitude of b is -3
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4 years ago

A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t

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Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

$f = f_{0}\frac{v + v_{o}}{v - v_{s}}$

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer

v: is the speed of the sound = 343 m/s

$v_{o}$ : is the speed of the observer = 0 (it is heard in the town)

$v_{s}$ : is the speed of the source =?

The frequency of the train before slowing down is given by:

$f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}$ (1)

Now, the frequency of the train after slowing down is:

$f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}$ (2)

Dividing equation (1) by (2) we have:

$\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}$

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}$ (3)

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

$v_{s_{b}} = 2v_{s_{a}}$ (4)

Now, by entering equation (4) into (3) we have:

$\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}$

$\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}$

By solving the above equation for $v_{s_{a}}$ we can find the speed of the train after slowing down:

$v_{s_{a}} = 11.06 m/s$

Finally, the speed of the train before slowing down is:

$v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s$

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

I hope it helps you!

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3 years ago