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Ulleksa [173]
3 years ago
12

Wave-particle duality tells us that wave and particle models apply to all objects whatever the size, so why don't we observe wav

e properties in macroscopic objects?
Physics
1 answer:
Genrish500 [490]3 years ago
4 0

Answer:

Because the wavelengths of macroscopic objects are too short for them to be detectable.

Explanation:

Wavelength of an object is given by de Broglie wavelength as:

\lambda=\frac{h}{mv}

Where, 'h' is Planck's constant, 'm' is mass of object and 'v' is its velocity.

So, for macroscopic objects, the mass is very large compared to microscopic objects. As we can observe from the above formula, there is an inverse relationship between the mass and wavelength of the object.

So, for vary larger masses, the wavelength would be too short and one will find it undetectable. Therefore, we don't observe wave properties in macroscopic objects.

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An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to
AURORKA [14]

Answer:

Explanation:

Expression for relative velocity

= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

= (.54 + .82 )c/ 1+ \frac{.54 \times.82}{1}

= 1.36 c / 1.4428

= .94 c

β = .94

6 0
3 years ago
Dr. Matthews has submitted a proposal to the institutional review board (IRB) of a university. At this university, she intends t
OlgaM077 [116]
The IRB at the university will decide whether her study meets ethical guidelines before it is initiated. The importance of these codes of conduct is to safeguard research participants, the status of psychology and the researchers or psychologists themselves. Moral issues hardly yield a simple, unequivocal, right or wrong answer. It is consequently often a matter of judgment whether the research is justified or not. For instance, it might be that a study roots psychological or physical uneasiness to participants, maybe they agonize pain or maybe even come to solemn harm.
6 0
3 years ago
SOMEONE PLEASE HELP!!
Arisa [49]

Hello!

\large\boxed{\text{C. 7,350,000 J}}

Use the equation:

PE = mgh

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

7 0
2 years ago
A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of
Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0
3 years ago
A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
sladkih [1.3K]

Explanation:

Given that,

Charge 1, q_1=6.75\ nC=6.75 \times 10^{-9}\ C

Charge 2, q_2=4.46\ nC=4.46\times 10^{-9}\ C

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

F=\dfrac{kq_1q_2}{r^2}

F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}

F=6.84\times 10^{-8}\ N

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

5 0
3 years ago
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