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Fofino [41]
3 years ago
13

Explain quantum zeno effect in one sentence

Physics
1 answer:
otez555 [7]3 years ago
5 0
It’s a feature of quantum-mechanical systems allowing a particle's time evolution to be arrested by measuring it frequently enough with respect to some chosen measurement setting.
You might be interested in
a car travelling at 50km/h from rest covers a distance of 10km in 40minutes. Calculate the acceleration​
Margarita [4]

Answer:

9.67\cdot 10^{-3}m/s^2

Explanation:

We can solve the problem by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

For the car in this problem:

u = 0 (it starts from rest)

v=50 km/h \cdot \frac{1000}{3600}=13.9 m/s is the final velocity

s = 10 km = 10 000 m is the displacement

Solving for a, we find:

a=\frac{v^2-u^2}{2s}=\frac{13.9^2}{2(10000)}=9.67\cdot 10^{-3}m/s^2

7 0
3 years ago
Adding a best-fit line to the scatter plot shown below would be an example of _____.
baherus [9]
I think It would be C. Checking a prediction. Sorry if I’m wrong
3 0
3 years ago
Read 2 more answers
(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is
Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H  = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = \dfrac{1}{2}mv^2

Using Conservation of energy

KE_i + PE_i + KE_f + PE_f

0 + m g H = \dfrac{1}{2}mv^2 + 0

v = \sqrt{2gH}

v = \sqrt{2\times 9.8 \times 105}

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0
3 years ago
The generation of a magnetic field by an electric current is___ .
Alex787 [66]

Answer:

Electromagnetic induction

Explanation:

The process of generating electric current with a magnetic field. It occurs whenever a magnetic field and an electric conductor move relative to one another so the conductor crosses lines of force in the magnetic field.

5 0
2 years ago
What is the period (in hours) of a satellite circling Mars 100 km above the planet's surface? The mass of Mars is 6.42 × 1023 kg
scZoUnD [109]

To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as

F_c = F_g

\frac{mv^2}{r} = \frac{GmM}{r^2}

Where,

m = Mass

G =Gravitational Universal Constant

M = Mass of the Planet

r = Distance/Radius

Re-arrange to find the velocity we have,

v^2 = \frac{GM}{r}

At the same time we know that the period is equivalent in terms of the linear velocity to,

T = \frac{2\pi}{\frac{v}{r}}

v = \frac{2\pi r}{T}

If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,

v^2 = \frac{GM}{r}

( \frac{2\pi r}{T})^2 =  \frac{GM}{r}

T = \sqrt{\frac{4\pi^2 r^3}{GM}}

Replacing we have,

T = \sqrt{\frac{4\pi^2 (3500*10^3)^3}{(6.67430*10^{-11})(6.42*10^23)}}

T = 6285.09s (\frac{1min}{60s})(\frac{1hour}{60min})

T= 1.74hour

Therefore the correct answer is C.

7 0
3 years ago
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