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3 years ago

What force is left out of the Quantum Mechanics theory?

Physics

1 answer:

nirvana33 [79]3 years ago

4 0

Answer:

Quantum mechanics is a key hypothesis in material science that gives a portrayal of the actual properties of nature at the size of iotas and subatomic particles. It is the establishment of all quantum physical science including quantum science, quantum field hypothesis, quantum innovation, and quantum data science.

Explanation:

It is the greatest of issues, it is the littlest of issues. At present physicists have two separate rule books clarifying how nature functions. There is general relativity, which perfectly represents gravity and everything it overwhelms: circling planets, impacting worlds, the elements of the growing universe all in all. That is enormous. At that point there is quantum mechanics, which handles the other three powers – electromagnetism and the two atomic powers. Quantum hypothesis is very proficient at portraying what happens when a uranium molecule rots, or when singular particles of light hit a sun based cell. That is little.

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the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located

vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

charge q = 31.7mC= 31.7×10⁻³C

distance r = 2.80mm

distance from midpoint to point A = 2.00mm

attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector E = 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

E=2{4.79×10¹³}×(0.574)(-x)

E = 2×2.75 ×10¹³N/C(-x direction)

Vector E= 5.5 ×10¹³N/C(-x direction)

3 0

3 years ago

A ball with a mass of 170 g which contains 3.80×108 excess electrons is dropped into a vertical shaft with a height of 145 m . A

yaroslaw [1]

Answer:

A. F=6.65*10^{-10}N

B. south - north

Explanation:

A) We use the Lorentz force

F = qv X B

|F| = qvB

to calculate the magnitude of the force we need the speed of the of the ball.

$v_{f}^{2}=v_{0}^{2}+2gy\\v_{f}=\sqrt{0+2(9.8\frac{m}{s^{2}})(145m)}=53.31\frac{m}{s}$

and by replacing in the formula for the magnitude of the force we have (taking into account the excess of electrons)

$F=(3.8*10^{8})(1.602*10^{-19}C)(53.31\frac{m}{s})(0.205T)=6.65*10^{-10}N$

b. south - north (by the rigth hand rule)

I hope this is usefull for you

regards

8 0

3 years ago

A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget

mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

Explanation:

Given,

Mass, $m_{1}$ = 4 kg

Speed, $v_{1}$ = 5 m/s

$m_{2}$ = 1 kg

$v_{2}$ = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

$\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2$

By plugging in the values we get,

$KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\$

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

$KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)$

$KE = 40J + KE(lost)$

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0

3 years ago

Hi, I need some science help ASAP. Here's the question:

ch4aika [34]

Answer:

Mass of the box is 25kg

Explanation:

From Newton's second law of motion

Force= mass. acceleration

With the symbol

F=m.a

m=F/a m=(50N)/(2m/s²)

m=25kg

7 0

2 years ago

Why does sodium chloride and magnesium chloride conduct electricity?

mihalych1998 [28]

They only conduct when they are in solution form, because then their ions become mobile, and the ions conduct electricity.

Hope this helps!

5 0

2 years ago