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lisov135 [29]
4 years ago
13

Help quick

Physics
2 answers:
Colt1911 [192]4 years ago
4 0
Its not intermost electrons its D
Sergeeva-Olga [200]4 years ago
3 0
A the intermost electrons
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An unstrained horizontal spring has a length of 0.40 m and a spring constant of 340 N/m. Two small charged objects are attached
Hatshy [7]

Answer:

(a) Both the charges are positive or negative.

(b) Teh value of each charge is 1.53 x 10^-5 C.

Explanation:

Spring constant, K = 340 N/m

Natural length, L = 0.4 m

stretch, y = 0.033 m

(a) Let the charge on each sphere is q and they repel each other so the nature of charge of either sphere may be both positive or both negative.  

(b) The electrostatic force is balanced by the spring force.

\frac{kq^2}{(L + y)^2}=Ky\\\\\\\frac{9\times 10^9 q^2}{(0.4 +0.033)^2} = 340\times0.033\\\\q= 1.53\times 10^{-5} C

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3 years ago
A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. T
lukranit [14]

Answer:

(a) The period is 4s

(b) The angular frequency is pi/2 radians

(c) The amplitude is 0.37cm.

(d) The displacement at time is  (0.37 cm) cos((pi/2)*t)

(e) The Velocity at time t is v = (0.58 cm)(sin((pi/2)*t)

(f) The maximum speed is  v_{m} = -0.58 cm/s

(g) The  maximum acceleration is 0.91 cm/s^2

Explanation:

We have a particle which oscillates with frequency of f = 0.25 Hz about the point x = 0.At t = 0, the displacement of the particle is = 0.37 cm and its velocity is zero.

(a) The period of the oscillations is,

T = 1/f

so

T = 1/(0.25 Hz)

T = 4.0s

(b) The angular frequency is,

f = 2\pi f\\

f = = 2(\pi)(0.25 Hz) =\pi /2  \\ radians

(c) Since

The amplitude is the maximum displacement that the particle makes from the equilibrium point, or when the speed of the particle is zero,

that is

x_{m}= 0.37 cm

(d) The displacement as a function of t is given be,

x = x_{m} cos(ωt+Φ)

as x = x_{m  t = 0, we get cos(Ф) = 1 = 0

so this equation becomes

x= (0.37 cm) cos((pi/2)*t)

(e) Now we need to find the speed of the particle as a function of t

the speed is the derivative of the displacement that is

v = dx/dt = -(0.37)(pi/2)(sin((pi/2)*t)

so the velocity at time t is

 v = (0.58 cm)(sin((pi/2)*t)

(f) Since

v = v_{m} sin(ωt+Ф)

then from part (e) we get

v_{m} = -0.58 cm/s

(g)

The amplitude of the maximum acceleration is

a_{m} = x_{m ω^2

      = (0.37 cm) (pi/2) = 0.91 cm/s^2

this is the maximum acceleration

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