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Mice21 [21]
3 years ago
5

A proton is located at <3 x 10^-10> m. What is r, the vector from the origin to the location of the proton

Physics
1 answer:
Eduardwww [97]3 years ago
4 0

Complete Question

A proton is located at <3 x 10^{-10}, -5*10^{-10}  , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton

Answer:

The  vector position is   \=r=

Explanation:

From the question we are told that

  The  position of the proton is m

Generally the vector location of the proton is mathematically represented as

       \= r =

So substituting values

     \=r=

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Two identical positive charges exert a repulsive force of 6.1 × 10−9 n when separated by a distance 3.6 × 10−10 m. calculate the
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A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done o
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Answer:

0 J

Explanation:

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2 years ago
A truck traveling at a velocity of 33m/s comes to a halt by decelerating at 11m/s^2. How far does the truck travel in the proces
snow_lady [41]

Answer:

<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>

  • <u><em>Initial velocity = 33 m/s</em></u>
  • <u><em>Final velocity = 0 m/s</em></u>
  • <u><em>Average velocity = (33 + 0) / 2  m/s = 16.5 m/s</em></u>

Explanation:

  1. <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
  1. <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
  1. <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>
4 0
3 years ago
0/2 File Limit
slamgirl [31]

Answer:

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

y = v_y t + \frac{1}{2} at^2

so we have

-100 = 15 t - \frac{1}{2}(-9.81) t^2

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

8 0
3 years ago
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