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Sergeeva-Olga [200]
3 years ago
15

What changes in the original atom are expected as a result of this natural phenomenon

Physics
1 answer:
castortr0y [4]3 years ago
3 0
The answer is the atomic number and the mass number will decrease.
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Strong solar winds blew dust and gas out of the solar system during which phase of the development of the Sun?
ipn [44]
The choices can be found elsewhere and as follows:

a. <span>Alpha Centauri </span>
<span>c. </span><span>T-tauri </span>
<span>b. </span><span>The Big Bang </span>
<span>d. </span><span>Nebular
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I believe the correct answer from the choices listed above is option D. <span>Strong solar winds blew dust and gas out of the solar system during Nebular phase. This seems to be the most logical option from the choices. Hope this helps. Have a nice day.</span>
8 0
2 years ago
Read 2 more answers
If the input work on a machine is equal to it's output work. the machine has _____ efficiency.
AleksAgata [21]

Answer:

100% efficiency

Explanation:

the machine has _100%_ efficiency.

(fill in the blank)

7 0
3 years ago
What does Pascal's Law state? what are the implications of Pascal's law?
dybincka [34]

Answer:

Pascal's Law states that the pressure applied to a fluid in a closed container is transmitted equally to all points in the fluid and act in all directions of the container. ... Therefore, it can rightly be said that since the liquid does not flow, it definitely has equal pressure acting on it at all the points.

Explanation:

plzzzzzzz Mark my answer in brainlist

4 0
3 years ago
Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
DiKsa [7]

Answer:

a.\rm -1.49\ m/s^2.

b. \rm 50.49\ m.

Explanation:

<u>Given:</u>

  • Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).

At time t = 3 seconds,

\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

For the time interval of 2 seconds,

\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.

It is the displacement of the particle in 2 seconds.

7 0
3 years ago
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