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kenny6666 [7]
3 years ago
8

For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta

tement may be true.) Select all that are True. At the lowest point in the oscillation, the momentum is zero. At the lowest point in the oscillation, mg>kssmg>kss At the lowest point in the oscillation, mg
Physics
1 answer:
omeli [17]3 years ago
6 0

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, mg < ks_s

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

P = mv

P = 0

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

F_{spring} > mg

Ks > mg

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A point charge q1=2.0μC is located on the positive y axis at y=0.30m, and an identical charge q2 is at the origin. Find the magn
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Answer:

(A) 0.279N at angle 38.02°

(B) 0.701N

(C) 14.19°

Explanation:

(A) The net force on q3 is given as:

F = Fxi + Fyj

Fx is the x component of the force

Fy is the y component of the force

Fx = -F(1, 3)cos(90 - x) + F(2, 3)cos0

Fy = -F(2, 3)cosx - F(2, 3)cos90 = -F(2, 3)cosx

First let us find y and angle x from the diagram.

Using Pythagoras theorem,

y² = 0.3² + 0.4²

y² = 0.25

y = 0.5m

Using SOHCAHTOA to find x,

sinx = 0.4/0.5

x = 53.13°

Electrostatic force, F is given as:

F = kqQ/r²

Where k = Coulumbs constant

F(1,3) = (k*q1*q3) / r²

F(1, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.5²)

F(1, 3) = 0.288N

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * 2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = 0.45N

Therefore,

Fx = -0.288cos36.87 + 0.45

Fx = 0.22N

Fy = 0.288cos53.13

Fy = 0.172N

=> F = 0.22i + 0.172j

The magnitude of the force will be

F(mag) = √(0.22² + 0.172²)

F(mag) = 0.279N

The direction of the force makes will be

tanθ = Fy/Fx

tanθ = 0.172/0.22 = 0.781

θ = 38.02° to the x axis.

(B) q2 = - 2.0 * 10^(-6)

This implies that:

F(2,3) = (k*q2*q3) / r²

F(2, 3) = (9 * 10^9 * -2.0 * 10^(-6) * 4.0 * 10^(-6)) / (0.4²)

F(2, 3) = -0.45N

Therefore,

Fx = -0.288cos36.87 - 0.45

Fx = -0.68N

Fy = 0.172N

=> F = - 0.68i + 0.172j

The magnitude of the force will be

F(mag) = √((-0.68)² + 0.172²)

F(mag) = 0.701N

(C) The direction of the force makes will be

tanθ = 0.172/0.68

θ = 14.19° to the x axis

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