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2 years ago

6

A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.0062

9 T and the proton's velocity makes an angle of 137 ∘ with the field, what is the magnitude of the magnetic force acting on the proton?

1 answer:

galina1969 [7]2 years ago

4 0

Answer:

Force on the proton will be $.73\times 10^{-14}N$

Explanation:

We have given speed of proton is 20.3% of speed of light

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

So speed of proton $v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec$

Magnetic field B = 0.00629 T

Charge on proton $q=1.6\times 10^{-16}C$

Angle between velocity and magnetic field $\Theta =137^{\circ}$

Force on the proton is equal to $F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N$

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Alisiya [41]

Answer:

The workdone is $W = 9.28 * 10^{3} J$

Explanation:

From the question we are told that

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The density of the cylinder is $\rho = 0.346 * \rho_w$

Where $\rho_w$ is the density of freshwater which has a constant value

$\rho_w = 1000 kg/m^3$

Now

Let the final height of the device under the water be $= h_f$

Let the initial volume underwater be $= V_n$

Let the initial height under water be $= h_i$

Let the final volume under water be $= V_f$

According to the rule of floatation

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This is mathematically represented as

$\rho_c g V_n = \rho_w gV_f$

$\rho_c A h = \rho A h_f$

So $\frac{0.346 \rho_w}{\rho_w} = \frac{h_f}{h}$

=> $\frac{h_f}{h_c} = 0.346$

Now the work done is mathematically represented as

$W = \int\limits^{h_f}_{h} {\rho_w g A (-h)} \, dh$

$= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.$

$= \frac{g A \rho}{2} [h^2 - h_f^2]$

$= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]$

Substituting values

$W = \frac{(9.8 ) (4.19) (10^3)}{2} (0.588)^2 (1 - 0.346)$

$W = 9.28 * 10^{3} J$

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2 years ago

A force of 20 N acts toward the east. Another force of 27 N acts at the same point toward the west. What is the magnitude and di

Jet001 [13]

Since we are working in one dimension (left right or East West), we don't need to worry about angles! It's just simply a matter of adding things up!

First list out all the forces and add negative (-ive) signs to each of the 'west' forces like this.

20 East + (-27 West) + ? = 10 East

so it's easy to see that 20 + (-27) = -7
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3 years ago

2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

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$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

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3 years ago