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Romashka [77]
2 years ago
11

In an elastic collision, _______ energy is conserved.

Physics
1 answer:
Nonamiya [84]2 years ago
4 0

Answer:

Kinetic energy and momentum are conserved.

Explanation:

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A spin balancer rotates the wheel of a car at 480 revolutions per minute. if the diameter of the wheel is 26 inches, what road s
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Part A:
To answer this item, we need to convert the given revolutions per minute in speed expressed as miles per hour. In order to do so, we need to use the dimensional analysis and the appropriate conversion factor.
 
We calculate the circumference that is covered by the wheel.
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where C is the circumference and D is the diameter.

Substitute the known values,
       C = π(26 in) = 81.68 inches

The speed tested is calculated below.
      s = (81.68 inches/rev)(480 rev/min)(1 mil/63360 inches)(60 min/1 hr)
            s = 37.127 mil/h

<em>ANSWER: 37.127 mil/h</em>

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Similar to what was done in Part A, the setting of the balancer is calculated below,
       setting = (80 mil/h)(1h/60 min)(63360 in/1 mil)(1 rev/81.68 in)
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A particle (mass = 2.0 mg, charge = −6.0 μC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It en
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The acceleration would be  a = 0.003* 6 =0.018\  m/s^2    

Explanation:

The objective of this solution is to obtain the acceleration of the particle

       Now looking at Newton law which is mathematically represented as

                       F = ma

  Where F is the force experience by a particle

              m is the mass of the particle

              a is the acceleration of the particle

  And also this force is equivalent to magnetic force in a magnetic field which is mathematically represented as

                    F = qvB

Where q is the charge of the particle

             v is the velocity of the  charge

             B is the magnetic field the charge is under it influence

  Now equating this two formulas

                   ma = qvB

 Making a the subject we have

                  a = \frac{qvB}{m}

In the question the direction of the is in the positive x-axis which is i hence the direction would be in the i direction

      So substituting  (2.0i +3.0j+4.0k)mT = (2.0i +3.0j+4.0k)*10^{-3}T for B

                    a = \frac{q}{m} * v (2.0i +3.0j +4.0k)*10^{-3}

      Substituting       3.0 Km /s = 3.0*10^{3}\ m/s  for v  and -6.0 \muC = -6.0*10^{-6} C for q

                     a = \frac{-6.0*10^{-6}}{2.0*10^{-3}} * 3.0*10^{3} *(2i+3j+4k) *10^{-3}

                       a = 0.003 * 3i(2i+3j+4k)

                      a = 0.003 *((3*2)i \ \cdot i \ +(3*3) i \ \cdot \ j  \ + (3*4)i \ \cdot \ k)

According to vector multiplication

                                             i \cdot i = j \cdot j = k\cdot k = 1\\\\and \ i\cdot j = i\cdot k  = 0

     So

               a = 0.003* 6 =0.018\  m/s^2          

     

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3 years ago
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