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Dmitry_Shevchenko [17]
3 years ago
5

NEED HELP NOW!!! I WILL BRAINLIEST.

Physics
1 answer:
Sergio039 [100]3 years ago
7 0

Answer:

1. Carbon-12

2. Nitrogen-13

3. Carbon-13

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As a mass on a spring moves farther from the equilibrium position, how do the velocity, acceleration, and force change
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Refer to the diagram shown below.

m =  the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A =  the amplitude ( the maximum distance) of the mass from the equilibrium
        position

The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω =  the circular frequency of the motion
T =  the period of the motion so that ω = (2π)/T

The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)

In the equilibrium position,
x is zero;
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At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.

In the graphs shown, it is assumed (for illustrative purposes) that
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Since his speed is constant,

B.

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If the distance between two charges is doubled, by what factor is the magnitude of the electric force changed? F_e final/F_e, in
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To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

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Here,

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As the distance is doubled so,

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\frac{F_{final}}{ F_{initial}} = \frac{1}{4}

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