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3 years ago

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A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of

0.00687 T 0.00687 T and the proton's velocity makes an angle of 101 ∘ 101∘ with the field, what is the magnitude of the magnetic force acting on the proton?

1 answer:

iogann1982 [59]3 years ago

4 0

Answer:

$8.64283\times 10^{-14}\ N$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

v = Velocity of proton = $0.267\times c$

c = Speed of light = $3\times 10^8\ m/s$

B = Magnetic field = 0.00687 T

$\theta$ = Angle = $101^{\circ}$

Magnetic force is given by

$F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N$

The magnetic force acting on the proton is $8.64283\times 10^{-14}\ N$

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A cart moving at 10 m/s is brought to a stop by the force plotted in the force-time graph shown here. Find the impulse and the a

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Mass = 8.8 kg

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<u>We are given a graph of Force vs. Time. Looking at the graph we can see that the Force acts approximately between the time interval from 1sec to 4sec. </u>

Newton's Second Law relates an object's acceleration as a function of both the object's mass and the applied net force on the object. It is expressed as:

$F=ma$ Eqn. (1)

where

$F$ : is the Net Force in Newtons ( $N$ )

$m$ : is the mass ( $kg$ )

$a$ : is the acceleration ( $m/s^2$ )

We also know that the acceleration is denoted by the velocity ( $v$ ) of an object as a function of time ( $t$ ) with

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Now substituting Eqn. (2) into Eqn. (1) we have

$F=m\frac{v}{t}\\ \\Ft=mv$ Eqn. (3)

However since in Eqn. (3) the time-variable is present, as a result the left hand side (i.e. $Ft$ is in fact the Impulse $J$ of the cart ), whilst the right hand side denotes the change in momentum of the cart, which by definition gives as the impulse. Also from the graph we can say that the Net Force is approximately ≈ $22N$ and $t=4 sec$ (thus just before the cut-off time of the force acting).

Thus to find the Impulse we have:

$J=Ft\\J=(22N)(4sec)\\J=88 kg m/s$

So the impulse of the cart is $J=88kg m/s$

Then, we know that the cart is moving at $v=10 m/s$ . Plugging in the values in Eqn. (3) we have:

$(22N)(4sec)=(10m/s)m\\\\88=10m\\\\m=\frac{88}{10}\\ \\m=8.8kg$

So the mass of the cart is $m=8.8kg$ .

8 0

3 years ago

The internuclear distance between two closest Ar atoms in solid argon is about 3.8 A. The polarizability of argon is 1.66e-30 m3

oee [108]

Answer:

83.72 K

Explanation:

$\alpha$ = Polarizability of argon = $1.66\times 10^{-30}\ m^3$

I = First ionization = 1521 kJ/mol

r = Distance between atoms = 3.8 A

R = Gas constant = 8.314 J/mol K

T = Boiling point

Potential energy due to dispersion of gas is given by

$P=-\frac{3}{4}\frac{\alpha^2I}{r^6}\\\Rightarrow P=-\frac{3}{4}\frac{(1.66\times 10^{-30})^2\times 1521\times 10^3}{(3.8\times 10^{-10})^6}\\\Rightarrow P=-1044.01\ J/mol$

Kinetic energy is given by

$K=\frac{3}{2}RT$

The potential and kinetic energy will balance each other

$P=\frac{3}{2}RT\\\Rightarrow 1.04401\times 10^{-33}=\frac{3}{2}RT\\\Rightarrow T=\frac{1044.01\times 2}{3\times 8.314}\\\Rightarrow T=83.72\ K$

The boiling point of argon is 83.72 K

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