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Gnoma [55]
3 years ago
5

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

ring'
PLEASE SHOW YOUR WORK
Physics
1 answer:
lys-0071 [83]3 years ago
3 0
The mass is m =  3kg

The acceleration is a =  2m/s/s

Force is equal to mass times acceleration

F = m*a = 3kg*2m/s/s = 6 N

This force is equal to the tension in the string

T = 6 N
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Answer:

15 meters

Explanation:

The inicial energy of the ball is just potencial energy, and its value is:

E = m * g * h = m * g * 20,

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In the moment that the ball strickes the ground, all potencial energy transformed into kinetic energy, and 25% of this energy is lost, so the total energy at this moment will be:

E' = 0.75 * E = 0.75 * m * g * 20 = 15*m*g

This kinetic energy will make the ball goes up again, and at the maximum height, all kinetic energy is transformed back into potencial energy.

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A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i
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Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive  x-axis  i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis  (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative  z -axis

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Given that I = 7.43 \frac{kW}{cm^2} =  7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}

Making E_{rms} the subject we have

                   E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }

Substituting values as given on the question

                E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }

                          = 2.37*10^5 \ V/m

The amplitude of the electric field is mathematically represented as

                  A_E = \sqrt{2} * E_{rms}

                         = \sqrt{2} * 2.37*10^5

                        A_E= 3.35*10^5 V/m

The amplitude of the magnetic field is mathematically represented as

                       A_M = \frac{A_E}{c}

Substituting value

                      A_M = \frac{3.35 *10^5}{3.0*10^8}

                             A_M= 1.12 *10^{-3} T

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