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3 years ago

According to quantum mechanics the motions of subatomic particles may be described as

Physics

1 answer:

SSSSS [86.1K]3 years ago

6 0

Answer:

Waves

Explanation:

Subatomic particles are particles that are of the subdivision of an atom and are smaller that atom, they are also described as waves in quantum mechanics.

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When an experiment is replicated, how should the results of the two experiments compare?

olasank [31]

When the experiment is replicated, this means the conduction of the second experiment, should be related, or similar, to your first results.

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What is a single cell organism able to do?

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A pipe open at both ends has a fundamental frequency of 220 Hz when the temperature is 0°C. (a) What is the length of the pipe?

Allushta [10]

Explanation:

It is given that,

Fundamental frequency, f = 220 Hz

(a) We know that at 0 degrees, the speed of sound in air is 331 m/s.

For open pipe, $\lambda=2l$

l is the length of pipe

Also,

$v=f\lambda$

$l=\dfrac{v}{2f}=\dfrac{331}{2\times 220}=0.75\ m$

(b) Let f' is the fundamental frequency of the pipe at 30 degrees and v' is its speed.

$v'=331\sqrt{1+\dfrac{T}{273}}$

$v'=331\sqrt{1+\dfrac{30}{273}}$

v' = 348.71 m/s

So, $f'=\dfrac{v'}{\lambda}$

$f'=\dfrac{348.71}{2\times 0.75}$

f' = 232.4 Hz

Hence, this is the required solution.

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3 years ago

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What energy transformation occurs when a skydiver first jumps from a plane?

SSSSS [86.1K]

Hi
I think the answer is:
GRAVITATIONAL POTENTIAL ENERGY TRANSFORMS INTO KINETIC ENERGY.
HOPE IT HELPS.

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A particle of mass 2.37 kg is subject to a force that is always pointed towards East or West but whose magnitude changes sinusoi

Pavlova-9 [17]

Answer:

$v(1.5)=0.7648\ m/s$

Explanation:

<u>Dynamics</u>

When a particle of mass m is subject to a net force F, it moves at an acceleration given by

$\displaystyle a=\frac{F}{m}$

The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by

$F=2cos1.1t$

The variable acceleration is calculated by:

$\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}$

$a=0.8439cos1.1t$

The instant velocity is the integral of the acceleration:

$\displaystyle v(t)=\int_{t_o}^{t_1}a.dt$

$\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt$

Integrating

$\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}$

$\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)$

$\boxed{v(1.5)=0.7648\ m/s}$

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3 years ago