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slega [8]
3 years ago
5

Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe

en them is 4N. What will the magnitude of this force become if the distance of separation were to double between them?
Physics
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

From the formula of force:

F =  \frac{kAB}{ {r}^{2} }  \\

since AB and k are constants:

F \:  \alpha  \:  \frac{1}{ {r}^{2} }  \\  \\ F =  \frac{x}{ {r}^{2} }

x is a constant of proportionality

• when force is 4N, separation distance is 1

4 =  \frac{x}{1}  \\ x = 4

therefore, equation becomes

F =  \frac{4}{ {r}^{2} }  \\

when r is doubled, r becomes 2. find F:

F =  \frac{4}{ {2}^{2} }  \\  \\ F =  \frac{4}{4}  \\  \\ { \underline{force \: is \: 1N}}

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What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field
earnstyle [38]

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

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3 0
3 years ago
To drive defensively means taking proactive measures to avoid accident situations regardless of their potential causes.
oksian1 [2.3K]

Answer:

True

Explanation:

defines defensive driving skills can be seen  as driving with a conscious mindset, in order  save lives, time, and money, in spite of the conditions around you and the actions of others.

It entails taking proactive measures such avoiding oncoming speeding vehicles, making sure you can see the rear tires of the vehicle in front of you in a traffic jam, always obeying traffic rules. This is done to protect the lives and property of both yourself and the people around you, regardless of he conditions that you are faced with while driving.

5 0
3 years ago
An 8.0-kg history textbook is placed on a 1.25-m high desk. How large is the gravitational potential energy of the textbook-Eart
denpristay [2]
<h2>Answer</h2>

Gravitational potential energy of the textbook-Earth system relative to the desk = 98 N/m

<h2>Explanation</h2>

Given that,

Mass of history book = 8kg

height of book from ground = 1.25m

acceleration due to gravity = 9.8m/s²

<h3>gravitational potential energy formula</h3><h2>GPE = Fg⋅h</h2><h2>GPE = mgh</h2>

where ,

m = mass of object(book)

g = acceleration due to gravity

h = the altitude of the object.

GPE = 8(9.8)(1.25)

       = 98 N/m

So, gravitational potential energy of the textbook-Earth system relative to the desk = 98 N/m

5 0
4 years ago
Read 2 more answers
A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3/2 km in 3.5 min. How fast (in m/s) is the car m
timama [110]

Answer:

7.9m/s

Explanation:

we need to convert the given values to metres and seconds

3/2km = 1500m and 3.5min = 210s

Using the first key equation of accelerated motion:

d = ((vi+vf)/2) x t

1500m = ((6.4+vf)/2) x 210 (plug in values)

3000m = (6.4 + vf) x 210 (get rid of the denominator by multiplying both sides by 2)

14.29 = 6.4 + vf (divide both sides to get rid of time)

vf = 7.885 m/s (subtract 6.4 from the right side to isolate vf)

2 significant digit answer is 7.9m/s

4 0
3 years ago
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givi [52]
Hello! The anwser is 2.7! I hope this helps! ^^
4 0
4 years ago
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