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padilas
3 months ago
15

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

the ramp exerts on the sack and the rate of increase in the speed of sack at this instant.
y
*13–84. the 2-lb block is released from rest at a and slides down along the smooth cylindrical surface. if the attached spring has a stiffness k = 2 lb> ft, determine its unstretched length so that it does not allow the block to leave the surface until u = 60°.
a
prob. 13–82
prob. 13–84
2,fuuu 2,frur
Physics
1 answer:
Softa [3K]3 months ago
4 0
The second question necessitates a figure to provide an answer. For the initial question
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s²
The ramp's reaction force is
F = 8 kg (5.63 m/s²)
F = 45 N
Differentiate the kinematic equation with respect to time to find the velocity's rate of increase.
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Keith_Richards [3271]

Answer:

The radius is r = 4.434 *10^{-5} \ m

Explanation:

The problem states that

    The magnetic field is  B = 90 mT = 90*10^{-3} \ T

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In general, for a collision to happen, the centripetal force on the electron in its orbit must equal the magnetic force acting on it  

   This can be mathematically expressed as

   \frac{mv^2}{r} = qvB

=>    r = \frac{m* v}{q * B}

Where  m denotes the electron’s mass, which has a value of m = 9.1 *10^{-31} \ kg  

             v signifies the escape velocity, mathematically represented as

                v = \sqrt{\frac{2 * KE}{m} }

Thus,

       r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }

     applying indices

    r = \frac{\sqrt{2 * KE * m} }{qB}

substituting these values

   

       

r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}

       

r = 4.434 *10^{-5} \ m

     

6 0
2 months ago
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Answer:

1.2 × 10^27 neutrons

Explanation:

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2 months ago
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