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3 years ago

How is a net electric charge produced

1 answer:

4 0

If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.

Source: https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/electric-charge-ap/a/electric-charge-ap1

Hope this helps :)

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Why does a black hole have a stronger gravitational pull than the star that collapse to form it?

Studentka2010 [4]

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

8 0

2 years ago

Two waves are superposed. one wave has an amplitude of 5 cm, and other has an amplitude of 4 cm. What is the resultant amplitude

Vsevolod [243]

If both waves have the same wavelength, then the amplitude of
their sum could be anything between 1 cm and 9 cm, depending
on the phase angle between them.

If the waves have different wavelengths, then the resultant is a beat
with an amplitude of 9 cm.

4 0

2 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

3 0

3 years ago

A small airplane with a wingspan of 18.0 m is flying due north at a speed of 63.6 m/s over a region where the vertical component

choli [55]

Answer:

(a) ε = 1373.8.

(b) The wingtip which is at higher potential.

Explanation:

(a) Finding the potential difference between the airplane wingtips.

Given the parameters

wingspan of the plane is = 18.0m

speed of the plane in north direction is = 70.0m/s

magnetic field of the earth is = 1.20μT

The potential difference is given as:

ε = Blv

where ε = potential difference of wingtips

B = magnetic field of earth

l = wingspan of airplane

v = speed of airplane

ε = 1.2 x 18.0 x 63.6

ε = 1373.8

(b) Which wingtip is at higher potential?

The wingtip which is at higher potential.

5 0

2 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

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4 0

1 year ago

How is a net electric charge produced​

How is a net electric charge produced