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1 month ago

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if m represents mass in kg, v represents speed in m/s, and r represents radius in m, show that the force F in the equation F=mv^

2 /r can be expressed in the unit kg×m/s^2

1 answer:

Keith_Richards [3.2K]1 month ago

4 0

This method is termed dimensional analysis, which focuses solely on the measurement units without considering their values. You need to perform calculations using variables while eliminating similar items from both the numerator and denominator. The outcome is as follows:

F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²].

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A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

serg [3582]

A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

$q_2 = Ne$

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

<pthe charge="" of="" a="" single="" electron="" is="" known="" to="" be="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" positive="" cork="" results="" from="" n="" excess="" electrons="" which="" can="" be="" depicted="" as="">

$q_1 = -Ne$

<pby calculating="" for="" n="" we="" derive="" the="" number="" of="" electrons="" cork="" has="" lost:="">

$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

</pby></pthe></pthe></pafter></pthe></pwe></pthus></pin></pmore></pas>

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1 month ago

A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size

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Answer:

The final size is nearly the same as the initial size because the increase in size $1.055\times 10^{- 7}$ is remarkably small

Solution:

According to the problem:

The proton beam energy is E = 250 GeV = $250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J$

Distance traveled by the photon, d = 1 km = 1000 m

Proton mass, $m_{p} = 1.67\times 10^{- 27} kg$

Initial size of the wave packet, $\Delta t_{o} = 1 mm = 1\times 10^{- 3} m$

Now,

This operates under relativistic principles

The rest mass energy for the proton is expressed as:

$E = m_{p}c^{2}$

$E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J$

This proton energy is $\simeq 250 GeV$

Thus, the speed of the proton, v $\simeq c$

The time to cover 1 km = 1000 m of distance is calculated as:

T = $\frac{1000}{v}$

T = $\frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s$

According to the dispersion factor;

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}$

$\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}$

Thus, the widening of the wave packet is relatively minor.

Hence, we can conclude that:

$\Delta t_{o} = \Delta t$

where

$\Delta t$ = final width

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