Answer:
Explanation:
This is going to sound like an absurd answer, but sometimes physics can be a little strange.
This answer is weird because of the definition of displacement. It means the distance from the starting point to the ending point, disregarding what happened in between. The point is that the astronaut is at the starting point of his orbit. By definition the starting and ending points are the same. His displacement is 0.
So the answer is you have the greater displacement when you walked one way to school. The starting point and the ending point are different. You have gone further.
However just to make things a little nasty, when you walk home again, your displacement will be the same as the astronaut's -- 0 meters because you will be right back where you started from.
The old electricity meters with disk using sapphire bearings for friction to be minimal to not aggravate metering
Answer:
electrons exist in specified energy levels
Explanation:
In its gold-foil scattering with alpha particles, Rutherford proved that the plum-pudding model of the atom theorised by Thomson was wrong.
From his experiment, Rutherford inferred that the atom actually consists of a very small nucleus, where all the positive charge is concentrated, and the rest of the atom is basically empty, with the electrons (negatively charged) orbiting around the nucleus at very large distance.
However, Rutherford did not specify anything about the orbits of the electrons. Later, Bohr predicted that the electrons actually orbit the nucleus in specific orbits, each orbit corresponding to a specific energy level. Bohr's model found confirmation in the observation of the emission spectrum lines: when an electron in one of the higher energy level jumps down into an orbit with lower energy, the atom emits a photon which has an energy exactly equal to the difference in energy between the two orbits (and this energy of the photon corresponds to a precise wavelength).
The force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.
<h3>How to calculate force?</h3>
The force needed to push an object can be calculated by multiplying the mass of the object by its acceleration as follows:
Force = mass × acceleration
According to this question, a car of mass 800 kg has an acceleration of 2.0 ms−². The force is calculated as follows:
Force = 800kg × 2m/s²
Force = 1600N
Therefore, the force needed to give a car of mass 800 kg an acceleration of 2.0 ms-² is 1600N.
Learn more about force at: brainly.com/question/13191643
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The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.
<span>In that particular situation, you can prove it like this: </span>
<span>initial velocity is Vo </span>
<span>launch angle is α </span>
<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>
<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>
<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>
<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>
<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>
<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>
<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>
