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4 years ago

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A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13

N, what is the strength of the magnetic field? (Express your answer using two significant figures.)

1 answer:

tia_tia [17]4 years ago

8 0

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

$F = qvB$

here we have

$B = \frac{F}{qv}$

here we know that

$F = 4.8 \times 10^{-13} N$

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

$v = 4 \times 10^6 m/s$

now from above equation we have

$B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}$

$B = 0.75 T$

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A 0.26 kg rod of length 80 cm is suspended by a frictionless pivot at one end. It is held horizontal and released.

Daniel [21]

Answer:

a) $a_{center} = 7.38 ~m/s^2$

b) $a_{end} = 14.77 ~m/s^2$

c) $v_{center} = 2.43~m/s$

Explanation:

a) Immediately after the rod is released, <u>the rod is still horizontal but now subject to gravity.</u> Since one end of the rod is fixed, then the weight of the rod applies a torque. Then by Newton's Second Law, the acceleration can be found.

$\tau =I\alpha$

where I is the moment of inertia of the rod with respect to its fixed end, and α is the angular acceleration.

The net torque of the rod is

$\vec{\tau} = \vec{r} \times \vec{F}\\\tau = rF\sin(90) = rF$

where r is the distance from center of the mass to the fixed end, so r = 0.4 m.

The weight of the rod is w = mg = 0.26 x 9.8 = 2.54 N.

So the net torque is τ = 1.01 Nm.

The moment of inertia of the rod is

$I = \frac{1}{3}mL^2 = \frac{1}{3}(0.26)(0.8)^2 = 0.055~kg m^2$

So, the Newton's Second Law yields

$\tau = I\alpha\\\alpha = \frac{\tau}{I} = \frac{1.01}{0.055} = 18.47$

<u>The relation between angular acceleration and linear acceleration is a = αr </u>

So, the linear acceleration of the rod is

$a = \alpha r = 7.38~m/s^2$

b) Using the same relationship between angular acceleration and linear acceleration, the linear acceleration of the end of the rod can be found.

$a = \alpha L = 14.77~m/s^2$

c) The conservation of energy can be used to find the velocity when the rod is vertical.

$K_1 + U_1 = K_2 + U_2\\0 + mg(L/2) = \frac{1}{2}I\omega^2 + 0\\(0.26)(9.8)(0.4) = \frac{1}{2}(0.055)\omega^2\\\omega = 6.08~rad/s$

The linear velocity is v = ωr, so

v = 2.43 m/s.

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3 years ago

Four identical masses of 2.5 kg each are located at the corners of a square with 1.0-m sides. What is the net force on any one o

Ghella [55]

Answer:

F=8.0*10^{-10}N

Explanation:

See the attached file for the masses distributions

The force between two masses at distance r is expressed as

$F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\$

since the masses are of the same value, the above formula can be reduce to

$F=\frac{Gm^{2}}{r^{2} }\\$

using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is

$F_{12} =\frac{Gm^{2}}{r^{2} }j\\$

The force on the lower left corner of the mass due to the lower right side of the mass is

$F_{14} =\frac{Gm^{2}}{r^{2} }i\\$

The force on the lower left corner of the mass due to the upper right side of the mass is

$F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\$

The net force can be express as

$F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}$

if we insert values we arrive at

$F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j$

if we solve for the magnitude, we arrive at

$F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}$

Hence the net force on one of the masses is

$F=8.0*10^{-10}N$

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