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telo118 [61]
3 years ago
10

A different group is worried that the motion detector will not

Physics
1 answer:
nekit [7.7K]3 years ago
6 0

The energy is not  conserved because:

1. Energy may still be leaving the system through friction.

2. There could be measurement error that account for the missing energy

Given that a different group roll a round  object down a ramp instead of dropping it. They measure the initial  height to calculate gravitational potential energy, use the motion  detector to track the final velocity, and calculate the kinetic energy of  the object at the end.

Despite all their careful measurements, it appears that energy is not

conserved because

1. Energy may still be leaving the system through friction. In which the energy lost will be converted to thermal energy

2. There could be measurement error that account for the missing energy. The may occur due to systematic error or parallax error

The object is not moving as fast at the bottom as the initial gravitational potential would have led them to predict. This is not as a result of rotational energy. There is nothing like rotational energy.

Learn more here: brainly.com/question/23503524

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(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas
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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = \frac{k*q}{r}

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = \frac{k*q1}{x} +\frac{k*q2}{(1.63m-x)} = 0

⇒k*q1* (1.63m - x) = -k*q2*x:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

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Answer:

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