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Lemur [1.5K]
2 years ago
10

a central concept in quantum mechanics is that both matter and are alternate forms of the same entity and therefore both exhibit

dual characteristics of particles and of . this model allows a better understanding of the behavior of tiny particles such as electrons.
Physics
2 answers:
Stels [109]2 years ago
4 0

In quantum mechanics, a central concept is that both matter and <u>energy</u> are alternate forms of the same entity and therefore both exhibit dual characteristics of particles and of <u>waves</u>.

Matter can be defined as anything that has mass and is able to occupy space.

Thus, any physical object or substance that is found on Earth is typically composed of matter.

Similarly, energy is highly affected by the mass of a any physical object or substance just like matter,

Hence, both energy and matter are known to be made up of atoms and as a result of this fact, exhibit dual characteristics of particles and of waves.

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In conclusion, this central concept makes it easier for us to better understand the behavior of tiny particles such as electrons.

Find more information: brainly.com/question/17203857

kkurt [141]2 years ago
4 0

Answer:

energy, waves

Explanation:

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Ms. Mary Mack walked around her block from her house for 200 meters. She arrived back at her house in 15 minutes. What was her d
inysia [295]

Answer:

d = 0 [m]

Explanation:

Displacement is understood as the length and direction that a body travels to move from an initial point to an endpoint.

This displacement is represented with a vector or straight line that indicates the distance of the displacement and its length.

This displacement in an easier way to understand. It is the distance between the start point and the endpoint of the journey. Since the second point is equal to the first point, since Mary returns to the same place, there is no difference between the displacement.

Therefore the displacement is zero.

5 0
3 years ago
A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that
lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0
2 years ago
A 0.400-kg ice puck, moving east with a speed of 5.86 m/s , has a head-on collision with a 0.900-kg puck initially at rest.
andreev551 [17]

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}

Energy

\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})

m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}

Where:

m_{1}, m_{2} - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

v_{1,o}, v_{2,o} - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

v_{1}, v_{2} - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If m_{1} = 0.400\,kg, m_{2} = 0.900\,kg, v_{1,o} = +5.86\,\frac{m}{s}, v_{2,o} = 0\,\frac{m}{s}, the system of equation is simplified as follows:

2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}

13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}

Let is clear v_{1,f} in first equation:

0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}

v_{1,f} = 5.86-2.25\cdot v_{2,f}

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}

13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}

13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}

2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0

2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0

There are two solutions:

v_{2,f} = 0\,\frac{m}{s} or v_{2,f} = 3.606\,\frac{m}{s}

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: (v_{2,f} = 3.606\,\frac{m}{s})

v_{1,f} = 5.86-2.25\cdot (3.606)

v_{1,f} = -2.254\,\frac{m}{s}

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

3 0
3 years ago
In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m
Serga [27]

Answer:

q = 6.48 \times 10^{-14} C

Explanation:

Deflection in the drop is due to electric field force

so we will have

F = qE

acceleration of the drop is given as

a = \frac{qE}{m}

a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}

a = 7.75 \times 10^{15} q

now we know that time to cross the plates is given as

t = \frac{D}{v}

t = \frac{0.02}{18}

t = 1.11 \times 10^{-3} s

now the deflection is given as

d = \frac{1}{2}at^2

0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2

0.310 \times 10^{-3} = 4.78 \times 10^9 q

q = 6.48 \times 10^{-14} C

5 0
2 years ago
Two electric charges, held a distance, dd, apart experience an electric force of magnitude, FF, between them. If one of the char
lorasvet [3.4K]

Answer:

F'=2F

Explanation:

The Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them:

F=\frac{kq_1q_2}{d^2}

In this case, we have q_1'=2q_1:

F'=\frac{kq'_1q_2}{d^2}\\F'=\frac{k(2q_1)q_2}{d^2}\\F'=2\frac{kq_1q_2}{d^2}\\F'=2F

3 0
3 years ago
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