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Norma-Jean [14]

4 years ago

12

oppositely charged objects attract each other this attraction holds electrons in Adams and hold on to one another in many compou

nds however ernest rutherfords model of the atoms failed to explain why electrons were not pulled into the aromic nicleus by this attraction

1 answer:

gavmur [86]4 years ago

5 0

Yes. Newton's laws do not apply at the subatomic level. There you are entering the weird world of Quantum Theory. Go read about the '2 holes experiment'. It will blow your mind.

You might be interested in

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

$h = 0.50 - 0.03 = 0.47m$

now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

$v = 3.03 m/s$

7 0

4 years ago

A two-turn circular wire loop of radius 0.63 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.219 T. I

Basile [38]

Answer:

The magnitude of the average induced emf in the wire during this time is 9.533 V.

Explanation:

Given that,

Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

$E=-\dfrac{d\phi}{dt}$

$E=-B\dfrac{dA}{dt}$

$E=-B\dfrac{A_{2}-A_{1}}{dt}$

$E=B\dfrac{A_{1}-A_{2}}{dt}$ .....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

$l=2\times 2\pi\times r_{1}$

$l=2\times 2\pi\times 0.63$

$l=7.916\ m$

The length when wire is in one loop

$l=2\pi\times r_{2}$

$7.916=2\times \pi\times r_{2}$

$r_{2}=\dfrac{7.916}{2\times \pi}$

$r_{2}=1.259\ m$

We need to calculate the initial area

$A_{1}=N\times\pi\times r_{1}^2$

Put the value into the formula

$A_{1}=2\times3.14\times(0.63)^2$

$A_{1}=2.49\ m^2$

The final area is

$A_{2}=N\times\pi\times r_{2}^2$

$A_{2}=1\times\pi\times(1.259)^2$

$A_{2}=4.98\ m^2$

Put the value of initial area and final area in the equation (I)

$E=0.219\dfrac{2.49-4.98}{0.0572}$

$E=-9.533\ V$

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

6 0

3 years ago

To make the jump, Neo and Morpheus have pushed against their respective launch points with their legs applying a _____ to the la

Nimfa-mama [501]

Force is the interaction that changes the motion of an object. If we want to make a jump, our legs have to exert a force to the ground (or launch points), so that we can gain an equal and opposite force that changes our motion from being static on the ground to moving forward.

6 0

3 years ago

Read 2 more answers

What happens when a particle of matter meets its corresponding antiparticle of antimatter?

shutvik [7]

Answer:

The combined mass of the two particles is completely transformed into energy (photons). This process is called matter-antimatter annihilation.

Explanation:

6 0

3 years ago

Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x

GalinKa [24]

Answer:

a) (-367231.63i , 367231.63i, 0) N/C

b) (0 , 0 , 367231.63i ) N/C

Explanation:

a)

Case x < -2.15

$E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i$

Case x > 2.15

$E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i$

Case -2.15 < x <+2.15

$E =E_{+} - E_{+} \\\\E = 0$

b)

Case x < -2.15

$E =E_{+} - E_{+} \\\\E = 0$

Case x > 2.15

$E =E_{+} - E_{+} \\\\E = 0$

Case -2.15 < x <+2.15

$E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i$

7 0

3 years ago