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STatiana [176]
3 years ago
15

Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3_m). It has a positive char

ge that is three times the magnitude (3_(qe)) of the charge on an electron. What is the ratio of the speed v_c that the cyberon would have when it reaches the upper plate after being released from rest at position h_0 to the speed ve that the electron would have?
Physics
1 answer:
Gre4nikov [31]3 years ago
3 0

Answer:

The answer is "The last choice".

Explanation:

Please find the complete question in the attachment.

In an external electric field, its electrical energy at positive charge becomes directed to just the electrical domain. Therefore it will speed towards its base plate whenever cyber one is released to rest at h0. It was never going to reach the top plate. Thus,  the last choice corrects because in this the cyber-on never reaches its upper stage.

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A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
Nady [450]

Answer:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2\pi  √m/ k

a) empty car

           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

           k x + kx + kx + kx = k_eq x

           k_eq = 4 k

           k = k_eq / 4

           k = 12 929.18 / 4

            k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

angular velocity and frequency are related

              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

           

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

3 0
3 years ago
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