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3 years ago

Two objects, each with a charge of +0.15 C are separated by a distance of 3 meters.

Physics

1 answer:

Mrac [35]3 years ago

6 0

Answer:

A . Are the masses repelling or attracting

answer:

So, positive energy densities ("masses") attract each other by gravitational interaction. This is the general idea.

B. What is the magnitude of the electrical force between the objects?

answer:

 The magnitude of the electrostatic force F between two point charges q1 and q2 is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

C. What would the magnitude of the force be if one of the charges was 1/3 the amount?

answer:

0.45

D. ) What would the force be if the distance between the charges was only 1 meter (with the original charges)?

answer;

Fmin=2.3×10^−28N

E: What would the force be if one of the charges was 1/3 the amount AND the distance was 1 meter?

answer;

Electrostatic force is directly related to the charge of each object. So if the charge of one object is doubled, then the force will become two times greater.

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We can calculate the length of each spring by using the relationship:
$F=kx$
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Re-arranging the equation, we have
$x= \frac{F}{k}$

The force applied to both spring is F=60 N. Spring A has spring constant of k=4 N/m, therefore its length with respect to its rest position is
$x_A= \frac{F}{k_A}= \frac{60 N}{4 N/m}=15 m$
Spring B has spring constant of k=5 N/m, so its length with respect to its rest position is
$x_B= \frac{F}{k}= \frac{60 N}{5 N/m}=12 m$

Therefore, the correct answer is
D.Spring A is 3 m longer than spring B because 15 – 12 = 3.

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3 years ago

Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5,

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Answer:

1) 5.52 cm , C) 5.5 cm

Explanation:

When a measurement is carried out, in addition to the value of the magnitude, the error or uncertainty of the measurement must occur, in a direct measurement with an instrument the uncertainty is equal to the appreciation of the instrument.

Uniform see the errors by the number of significant figures days, in this cases they are two decimals for which the appreciation of the instrument ± - 0.01

now we can analyze the measurements made

1) 5.52 cm. Validate. It is a valid measurement is within the uncertainty range

2) 6.63 cm. It does not validate. It is out of the error range

3) 5.5 cm Valid. It is within the given error range,

4) 5.93 cm Not Valid. It is out of the error range.

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