In the physical sciences, a particle (or corpuscule in older texts) is a small localized object to which can be ascribed several physical or chemical properties such as volume, density or mass. They vary greatly in size or quantity, from subatomic particles like the electron, to microscopic particles like atoms and molecules, to macroscopic particles like powders and other granular materials. Particles can also be used to create scientific models of even larger objects depending on their density, such as humans moving in a crowd or celestial bodies in motion.
The term 'particle' is rather general in meaning, and is refined as needed by various scientific fields. Anything that is composed of particles may be referred to as being particulate. However, the noun 'particulate' is most frequently used to refer to pollutants in the Earth's atmosphere, which are a suspension of unconnected particles, rather than a connected particle aggregation.
The energy-momentum tensor of a free particle with mass ##m## moving along its worldline ##x^\mu (\tau )## is
\begin{equation}
T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.
\end{equation}
Let contract...
The energy-momentum tensor of a free particle with mass ##m## moving along its worldline ##x^\mu (\tau )## is
\begin{equation}
T^{\mu\nu}(y^\sigma)=m\int d \tau \frac{\delta^{(4) }(y^\sigma-x^\sigma(\tau ))}{\sqrt{-g}}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}.\tag{2}
\end{equation}
The covariant...
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v = sqrt( (2 * charge of proton * 3000/e) / (mass of proton))
v = 1.893986024 x 10^`15
r = ( (mass of proton) * (velocity) ) / ((magnetic field) * (charge of proton))
r = 24715769.68 m
Anyone please help
m = 0.005
q = -70 x 10^-6 c
v = 30,000 m/s
Since there is no movement vertically Fb = Mg
So,
q . V . B = mg
So,
(70 x `10^-6) . (30,000) . B = (0.005) . (9.8)
So,
B = 0.0233333 or ~ 23 MT
F = ma
F = (6x10^-6) * 8
F = 4.8 * 10^-5
F = QBVsin(theta)
F/(BVsin(theta) = Q
(4.8 x 10^-5) / (5 x 10^-3) (4000) (sin(37)) = 3.98 x 10^-6 ~ 4 uc <---- THE RIGHT ANSWER IS -4 uc
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Hi.
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Homework Statement
From Griffiths GM 3rd p.266
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Homework Statement
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Homework Equations
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Homework Statement
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Greetings,
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Taken from the Wikipedia page:
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