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liraira [26]
3 years ago
14

When an object has a negative charge, which of these describes the electric field surrounding the object? A. It's neutral. B. It

gets stronger away from the object. C. It becomes positive away from the object. D. The electric field lines move into the object.

Physics
2 answers:
inysia [295]3 years ago
6 0

Answer:

<h2>D. The electric field lines move into the object.</h2>

Explanation:

It's important to know when an object is "charged", it means that object has an imbalance of charge. So, objects with more electrons than protons are charged negatively. Objects with more protons than electrons are charged positively.

So, in this case, an object has a negative charge, that means it has more electrons than protons.

Now, there's a concept called electric field lines which are imaginary lines to described an electrical field. According to Michael Faraday, an object charged negatively exerts electric field lines into itself, if the object has a positive charge, the electric field lines are from inside to outside the object.

Therefore, the right choice here is D.

The image attached shows an example of this.

Margaret [11]3 years ago
4 0
D. Electric field lines move into the object.
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IrinaVladis [17]
This is the same question that I just answered.

Have present the definition of acceleration:

         a = Δv / Δt, this is change in velocity per unit of time.

a and v are in bold to mean that they are vectors.

1) a body traveling in a straight line and increasing in speed: CORRECT:

Acceleration is the change in velocity, either magnitude or direction or both. So, a body increasing in speed is accelerated.

2) a body traveling in a straight line and decreasing in speed: CORRECT

A decrease in speed is a change in velocity, so it means acceleration.

3) a body traveling in a straight line at constant speed: FALSE.

That body is not changing either direction or speed so its motion is not accelerated but uniform.

4) a body standing still : FALSE.

That body is not changind either direction or speed.

5) a body traveling at a constant speed and changing direction: CORRECT.

The change in direction means that the body is accelerated. The acceleration due to change in direction is named centripetal acceleration.
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4 years ago
Why does the fed rarely change the reserve requirement
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The answer is:

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8 0
3 years ago
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Which form of Kepler’s third law can you use to relate the period T and radius r of a planet in our solar system as long as the
ser-zykov [4K]

T2=r In the form of Kepler's law that can use to relate the period T and radius of the planet in our solar systems

<u>Explanation:</u>

<u>Kepler's third law:</u>

  • Kepler's third law states that For all planets, the square of the orbital

     period (T) of a planet is proportional to the cube of the average orbital    radius (R).

  • In simple words T (square) is proportional to the R(cube) T²2 ∝1 R³3
  • T2 / R3 = constant = 4π ² /GM

      where G = 6.67 x 10-11 N-m2 /kg2

        M = mass of the foci body

8 0
4 years ago
The changing appearance of the Moon during the lunar cycle is the result of the changing -
julia-pushkina [17]

Answer:

relative positions of Earth, the Moon and the Sun.

Explanation:

A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.

Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).

Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.

There are three (3) types of lunar eclipse and these are;

1. Total lunar eclipse.

2. Partial lunar eclipse.

3. Penumbra lunar eclipse.

Hence, the changing appearance of the Moon during the lunar cycle is the result of the changing relative positions of Earth, the Moon and the Sun.

6 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
IgorC [24]

Answer:

The answers to the questions are;

a. The number of times the student run the flight of stairs to lose 1.00 kg of fat is 829.23 times.

b. The average power output, in watts and horsepower, as he runs up the stairs is 158.026 watts.

c. The act of climbing the stairs is not a practical way to lose weight has to lose 1 kg of fat, the student needs to workout for about 26.49 hrs or 1.104 days.

Explanation:

To solve the question, we write out the known variables as follows

1 g of fat = 9.00kcal

Number of steps the student climbs = 95 steps

Height of each step = 0.150 m

Time it takes for the student to reach the top of the stairs = 57.5 s.

Efficiency of human muscles = 20 %

Mass of student, m = 65 kg

a. From the question, the energy expended by the student in climbing the stairs is the "work done" by the student.

The "work done" is the height climbed resulting in the gaining of gravitational potential energy P. E..

That is work done, W, =  P. E. = m·g·h

Where:

h = The total height climbed by the student

g = Acceleration due to gravity = 9.81 m/s²

Therefore;

h = Height of each step × Number of steps the student climbs =

  = 0.150 m/(step) × 95 steps = 14.25 m

Therefore, P. E. = 65 kg × 9.81 m/s² × 14.25 m = 9086.5125 kg·m²/s²

                          = 9086.5125 J

We remember that the efficiency of the muscle is 20 %

The formula for efficiency is

Efficiency = \frac{Ene rgy Out put}{Energ y In put} \times 100 %

The work produced by the muscle =  Energy Output = 9086.5125 J

Energy input is given by

\frac{Out put} {Effici ency} = 9086.5125 J/ (0.2) = 45432.5625 J

= 45.432 kJ

From the question, 1 g of fat = 9.00 kcal and

1 kcal = 4186 J

Therefore 1 g of fat can release 9.00 kcal × 4186 J = 37674 J

Therefore 1 kg of fat = 1000 g = 1000 × 37674 J = 37674 kJ

To consume the energy in 1 kg of fat the student therefore will run up the foight of stairs \frac{37674 kJ}{45.432 kJ} times to make up the 37674 kJ energy contained in 1 kg of fat

That is  \frac{37674 kJ}{45.432 kJ} =  829.23 times

b. Power is the rate of doing work

That is Power output = \frac{ WorkO utput }{Time} = \frac{9086.5125 J}{57.5 s} = 158.026 watts

c. No as the activity student will have to spend a total time of

829.23 × 57.5 s = 47680.67 s climbing up the stairs alone  and

47680.67 s = ‪13.24 Hours climbing up of which if the time to climb down is the same s climbing up, then we ave total time = 2× ‪13.24 Hours  

= 26.49 hrs = 1.104 days exercising which is not humanly possible.

3 0
3 years ago
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