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Kryger [21]
3 years ago
11

Identify whether the following statements describe a change in acceleration. Explain your response

Physics
1 answer:
alekssr [168]3 years ago
5 0

Answer: The answer will be A

Explanation:

There is acceleration in all cases velocity changes. In options b and c the speed is constant, then there is no acceleration. In option a, the car was moving and then stop, therefore, its speed decrease to zero, and that represent an acceleration.

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Water is returned from earth’s surface to the atmosphere by
Tems11 [23]

Answer:

Evaporation

Explanation:

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4 0
3 years ago
Find the magnitude of the net electric force exerted on this charge. Express your answer in terms of some or all of the variable
Yuri [45]

Answer:

Th steps is as shown in the attachment

Explanation:

from the diagram, its indicates that twelve identical charges are distributed evenly on the circumference of the circle. assuming one of gthe charge is shifted to the centre of he circle alomng the x axis, as such the charge is unbalanced and there is need ot balanced all the identical charges for the net force to be equal to zero.

The mathematical interpretation is as shown in the attachment.

3 0
3 years ago
A non-relativistic particle of mass m moves in one dimension x under the force
marin [14]

Answer:

U = - (x⁴ / 4 - b x² / 2) , c) The function is zero  for x = 0 and √2

Explanation:

a and b) Strength and potential energy are related

               F = - dU / dx

Therefore to find the energy we must integrate

            ∫ dU = -∫ F dx

            ∫ dU = - ∫ (a x³ –b x) dx

Let's make the integration

             U = - (x⁴ / 4 - b x² / 2)

We evaluate the integral between the value

            U - U₀ = -x⁴ / 4 + x² / 2 - (-x₀⁴ / 4 + x₀² / 2)

The arbitrary constant is zero, so that U is zero in the zero position

                U₀ = 0 for x₀ = 0

c) Mechanical energy is the sum of kinetic energy plus potential energy

        Em = K + U

        Em = ½ m v² + ½ (x² -x⁴ / 2)

       E = ½ m v² + ½ x² (1 - x² / 2)

       Energy is positive

        2 (E –K) = x² (1-x² / 2)

At the return points K = 0

The zero points of this function are

     x = 0

     (1- x² / 2) = 0

     x² = 2

    x = √ 2

The function is zero

       x = 0 and √2

d) the movement is bounded for energy values ​​less than or equal to

            E <= ½ x² (1-x² / 2)

e) for this part we resolved Newton's second law

            F = m a

            ax³ - b x = m d²x / dt²

            d²x / dt² = -b / m x + a / m x³3

The linear term gives a simple harmonic movement

             w₀² = b / m

             d²x / dt² = - w₀² x + a / m x³

The frequencies are the frequencies of the harmonic movement plus a small change due to the non-harmonic part of the movement

6 0
4 years ago
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