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lukranit [14]
3 years ago
15

A 3.0-kg mass moving in the positive x direction with a speed of 10 m/s collides with a 6.0-kg mass initially at rest. after the

collision, the speed of the 3.0-kg mass is 8.0 m/s, and its velocity vector makes an angle of 35° with the positive x axis. what is the magnitude of the velocity of the 6.0-kg mass after the collision?
Physics
1 answer:
OverLord2011 [107]3 years ago
6 0
Sorry i dont know the answer 
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Using evidence from the article, defend the concept that Earth’s magnetic poles have swapped places over time.
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Answer: Scientists found evidence of Earths magnetic field reversal in rocks on the ocean floor at plate boundaries. These rocks have alternating polarity due to magnetization that occurred during their cooling period. Using radio metric dating, scientist estimate that reversals occur approximately every several hundred thousand years.

Explanation:

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If I turn on a light in a spaceship traveling 1C BACKWARDS, what happens to the photons? Speed -0-, or 1C in opposite direction?
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5 0
3 years ago
Why are the plates in a capacitor parallel to each other?
TiliK225 [7]
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6 0
3 years ago
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What is the difference between kinetic energy and potential energy​
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8 0
3 years ago
Initial velocity vector vA has a magnitude of 3.00 meters per second and points 20.0o north of east, while final velocity vector
garri49 [273]

Answer:

5.2\ \text{m/s}

70^{\circ} south of east

Explanation:

v_a = 3 m/s

\theta_a = 20^{\circ} north of east

v_b = 6 m/s

\theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

x and y component of v_a

v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

6 0
3 years ago
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