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klemol [59]
4 years ago
5

Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap

ply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.
Physics
2 answers:
eduard4 years ago
4 0

Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

Sloan [31]4 years ago
4 0

Complete Question:

An object oscillates back and forth on the end of a spring.

Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that apply. The object can have zero acceleration and, simultaneously, nonzero velocity. The object can have zero velocity and, simultaneously, nonzero acceleration. The object can have zero velocity and, simultaneously, zero acceleration. The object can have nonzero velocity and nonzero acceleration simultaneously.

Answer:

a) the object can have zero acceleration and, simultaneously, nonzero velocity

b) the object can have zero velocity and, simultaneously, nonzero acceleration

d) the object can have nonzero velocity and nonzero acceleration simultaneously

Explanation:

For an object oscillating back and forth on the end of a spring, when the object swings to the extremes, it momentarily stops and the velocity becomes zero. Rather than being zero, acceleration is maximum at these points because the net force at these extremes is maximum. This justifies option B

At the middle, which is the equilibrium position, the net force is zero because the object is motionless, and hence the acceleration is zero. At this point, the velocity is maximum and not zero. This justifies option A

When the object is swinging and it is neither at the middle nor at the extremes, both acceleration and velocity are not zero. This justifies option D

It is not possible for both the acceleration and the velocity of the swinging object to be simultaneously zero. Option C is wrong

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Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at
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Answer:

 E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

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For this exercise we will use that the electric field is a vector quantity, so the total field is

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bold font vectors .  We can work with the components of the electric field in each axis

X- axis

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the expression for the electric field is

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where r is the distance between the charge and the positive test charge

       

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

         E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

         E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

        E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

        E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

        E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

        E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

        E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

        E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

       E_{total x} = E_{13x} - E_{23x}

       E_{total x} = (3.516 - 1.23) 10⁶

       E_{total x} = 2.29 10⁶ N / A

       

       E_{total y} = -E_{13y} + E_{23y}

       E_{total y} = (-2.777 +1.86) 10⁶ N / A

       E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

         E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

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                E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

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                E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

                tan θ = E_total and / E_totalx

                θ = tan⁻¹ E_{total y} / E_{total x}

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Answer:

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