Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a
There would be very less percentage loss<span> of the kinetic energy during </span>the conversion<span> to internal energy, assuming that there is less air in the </span>surroundings<span>. Also, the friction will contribute to the conversion where if it is, the percentage loses is negligible.</span>
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
Answer:
<u><em>Definition of spectral line: </em></u><em>one of a series of linear images formed by a spectrograph or similar instrument and corresponding to a narrow portion of the spectrum of the radiation emitted or absorbed by a particular source.</em>
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<u><em>Definition of Wavelength:</em></u><em> can be defined as the distance between two successive crests or troughs of a wave. It is measured in the direction of the wave. ... Wavelength is inversely proportional to frequency. This means the longer the wavelength, lower the frequency.</em>
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<em>So, the spectrum is the range of wavelength in visible light. While, wavelength is the length of a wave.</em>
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Explanation:
I hope this helps!
Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
