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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.

#CARRYONLEARNING:)

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The amount of heat needed to increase the temperature of a solid sphere of diameter 2D of the same metal from 4°C to 7°C

aniked [119]

The amount of heat needed to increase the temperature of a solid sphere of diameter 2D of the same metal from 4°C to 7°C is is 8 times the initial amount of heat.

The temperature increment will lead to the increase in the internal energy of the object. This internal energy is the heat.

Given is the change in temperature ΔT = 7-4 =3°C., diameter D to 2D,

Q = Cp x ρ(4π/3)D³ x 3..................(1)

and Q' = Cp x ρ(4π/3)(2D)³ x 3

Q' = Cp x ρ(4π/3)8D³x 3..................(2)

Dividing both the equation, we have

Q' / Q =8

Q' = 8Q

Thus, the amount of heat needed to increase the temperature of a solid sphere of diameter 2D of the same metal from 4°C to 7°C is 8 times the initial amount of heat.

Learn more about heat.

brainly.com/question/1429452?

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5 0

2 years ago

Earth attracts a person with a gravitational force of 7.0 × 102 newtons. What is the magnitude of the force with which the indiv

Anuta_ua [19.1K]

This is a good time to review Newton's 3rd law of motion:
"For every action, there is an equal and opposite reaction."

Gravitational force always acts in pairs.
Whatever force the Earth attracts something with,
the thing attracts the Earth with exactly the same force.

If Earth attracts a person with a gravitational force of 7.0 × 10² newtons,
the person attracts Earth with a gravitational force of 7.0 × 10² newtons.

Your weight on Earth is the same as the Earth's weight on you !

5 0

3 years ago

bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4

serious [3.7K]

Answer:

Incomplete question: "A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?"

The normalized corrected value at 32.5 mm is 0.1638

Explanation:

The corrected light measurement at 29 mm is equal to:

20.7 - 5.1 = 15.6 mV

The corrected light measurement at 32.5 mm is equal to:

15.6 - 5.1 = 10.5 mV

To normalize the data to its maximum value means that the maximum value must be calculated and the data must be scaled using that value, as in this case the maximum value is 15.6 mm, then the normalized corrected value at 32.5 mm is equal to:

10.5 * 15.6 = 163.8 = 0.1638

6 0

3 years ago

Drag the tiles to the correct boxes to complete the pairs.

Dimas [21]

Brown dwarf is the first box
White dwarf is the second box
Black dwarf is the third box
Red giant is the fourth box
And
Black hole is the last box

5 0

3 years ago

13.An airplane starts from rest at the end of a runway and accelerates at a constant rate. In the first second, the airplane tra

Murrr4er [49]

Answer:

The answer is 3.33m

Explanation:

The acceleration "a" is constant.

Acceleration is the variation of velocity over time,

$\frac{dv}{dt} = a$ .

solving the last equation

$\int_{v_0}^v dv = a\int_0^t dt \rightarrow v-v_0 = at$ ,

where $v_0=0$ because the airplane starts from rest.

Once again, velocity is the variation of distance over time.

$\frac{dx}{dt} = at \rightarrow \int_{x_0}^x dx = a\int_0^t t\ dt$

then

$x- x_0 = \frac{1}{2}at^2$

where $x_0=0$ if we consider the end of the runway as the initial point (this step is for simplicity but you can let it expressed, it's going to cancel anyway).

If $x=1.11\ m$ at $t=1s$ , then

$a = \frac{2x}{t^2} = 2.22\ m/s^2$

and the final expression for the distance is

$x = 1.11 t^2$ .

If t = 2s, x = 4.44 m. Which means thad the additional distance is

$x(2s) - x(1s) = 4.44 - 1.11 = 3.33\ m$

8 0

4 years ago