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maw [93]
3 years ago
12

A 2.13-kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 5.00 N

/m. The object is displaced 3.54 m to the right from its equilibrium position and then released, initiating simple harmonic motion. (a) What is the force (magnitude and direction) acting on the object 3.50 s after it is released
Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

17.54N in -x direction.

Explanation:

Amplitude (A) = 3.54m

Force constant (k) = 5N/m

Mass (m) = 2.13kg

Angular frequency ω = √(k/m)

ω = √(5/2.13)

ω = 1.53 rad/s

The force acting on the object F(t) = ?

F(t) = -mAω²cos(ωt)

F(t) = -2.13 * 3.54 * (1.53)² * cos (1.53 * 3.50)

F(t) = -17.65 * cos (5.355)

F(t) = -17.57N

The force is 17.57 in -x direction

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Answer:

V = 26.95 cm³

Explanation:

Density is given by the formula :

ρ = m÷V

Density = mass ÷ Volume

Given both density and mass we rearrange, substitute and solve for Volume :

Rearranging the equation to make Volume the subject :

ρ = m÷V

ρV = m

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Now substitute :

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Hope this helped and have a good day

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fomenos

Complete Question

The complete question is shown on the uploaded image

Answer:

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Explanation:

The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as

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where F is the force, q is the charge and E is the electric field

we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.

we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge

Now from the question we are being told that the external electric field is the direction of the positive x axis

Hence this field would drive the positive charge i.e the nuclei to the right.

In order to further explain let consider this

Generally the electric field is always radially outward  originating from a positive point charge and radially in toward a negative point charge.

what  this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.

According to Coulomb's law which states that unlike term attract while like terms repel, it means that  the electron would move to the left of the conductor rod   while the nuclei would move to the right side of the conductor rod.

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