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Flauer [41]
3 years ago
5

How to contact brainly

Physics
2 answers:
LekaFEV [45]3 years ago
7 0

Answer:

go and rate brainly you can get question them

nevsk [136]3 years ago
5 0

Answer:

Go to your play store and rate their app

If you have any problem u can even mention that while you are rating them

I am sorry if I said anything wrong

You might be interested in
What could be a possible explanation why tectonic plates do not all move in the same direction
antiseptic1488 [7]

Answer:

tectonic plates move in opposite direction

Tectonic plates are not necessarily moving in opposite directions, but if there directions are different they form plate boundaries. If they were moving in the same direction they would become the same plate. The more opposite the direction, the more obvious the fault.

Examples of plate boundaries:

Divergent: moving away from each other - mid-Atlantic ridge

Convergent: moving towards each other - the Himalayas (Indian plate subducting under the Eurasian plate forcing up the mountains)

Transform: sliding laterally along each other - San Andreas fault

6 0
3 years ago
(1) A positive charge +3 C is separated from another positive charge of +5 C by a distance of 7m. What is the magnitude of the e
Aneli [31]

1. The magnitude of the electric force between the two charges is 2.8×10⁹ N (Option B)

2. The net charge on the molecule is -8×10⁻¹⁹ C (Option D)

3. The magnitude of the force between the charges is 16000 N (Option C)

4. The correct statement is: A neutral object has equal numbers of protons and electrons. (Option C)

<h3>1. How to determine the force</h3>
  • Charge 1 (q₁) = +3 C
  • Charge 2 (q₂) = +5 C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 7 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 3 × 5) / (7)²

F = 2.8×10⁹ N

<h3>2. How to determine the net charge on the molecule</h3>
  • Electron = 223 electrons
  • Proton = 218 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 218 - 223

Charge = -5 electrons

But

1 electron = 1.6×10⁻¹⁹ C

Thus,

Net Charge = -5 × 1.6×10⁻¹⁹ C

Net Charge = -8×10⁻¹⁹ C

<h3>3. How to determine the force</h3>
  • Charge 1 (q₁) = 2×10⁻⁴ C
  • Charge 2 (q₂) = 8×10⁻⁴ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.3 m
  • Force (F) =?

F = Kq₁q₂ / r²

F = (9×10⁹ × 2×10⁻⁴ × 8×10⁻⁴) / (0.3)²

F = 16000 N

<h3>4. What is a neutral object?</h3>

A neutral object is an object having equal numbers of protons and electrons. For example, an object with 4 protons and 4 electrons is said to be neutral as illustrated below

  • Electron = 4 electrons
  • Proton = 4 protons
  • Net Charge =?

Charge = Proton - Electron

Charge = 4 - 4

Charge = 0 (neutral)

Thus, the correct statement about neutral object, given in the question is: A neutral object has equal numbers of protons and electrons (Option C)

Learn more about Coulomb's law:

brainly.com/question/506926

#SPJ1

6 0
2 years ago
Pls help 100 points plssssssss
natima [27]

Answer:

d

Explanation:

8 0
2 years ago
Read 2 more answers
An egg is dropped from a building that is 61 m high.
Allisa [31]

Answer:

Initial Velocity = 0 m/s

Final Velocity = 34.6 m/s

time = 3.5 s

Explanation:

The initial velocity must be zero since, the egg must be at rest initially, before dropping.

<u>Initial Velocity = 0 m/s</u>

Now, for time we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = Height = 61 m

Vi = Initial Velocity = 0 m/s

g = 9.8 m/s²

t =time = ?

Therefore,

61 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t² = (61 m)(2)/(9.8 m/s²)

t = √(12.45 s²)

<u>t = 3.5 s</u>

Now, for final velocity we will use 1st equation of motion:

Vf = Vi + gt

Vf = 0 m/s + (9.8 m/s²)(3.5 s)

Vf = 34.6 m/s

3 0
3 years ago
A car with speed v and an identical car with speed 2v both travel the same circular section of an unbanked road. If the friction
yawa3891 [41]

Answer:

F'=\dfrac{F}{4}

Explanation:

Let m is the mass of both cars. The first car is moving with speed v and the other car is moving with speed 2v. The only force acting on both cars is the centripetal force.

For faster car on the road,

F=\dfrac{mv^2}{r}

v = 2v

F=\dfrac{m(2v)^2}{r}

F=4\dfrac{m(v)^2}{r}..........(1)

For the slower car on the road,

F'=\dfrac{mv^2}{r}............(2)

Equation (1) becomes,

F=4F'

F'=\dfrac{F}{4}

So, the frictional force required to keep the slower car on the road without skidding is one fourth of the faster car.

8 0
4 years ago
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