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3 years ago

An electron is accelerated through 1.90 103 V from rest and then enters a uniform 1.80-T magnetic field.

Physics

1 answer:

cluponka [151]3 years ago

3 0

Answer:

https://www.slader.com/discussion/question/an-electron-is-accelerated-through-240-times-103-v-from-rest-and-then-enters-a-uniform-170-t-magnetic-field-what-are-a-the-maximum-and-b-the-9e425fbd/

( Here is solution)

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A place or object used for comparison to determine if something is moving

castortr0y [4]

A reference point is needed to determine if something is in motion.

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3 years ago

3 What is the displacement of a satellite when it makes a complete round along its circular path?

kirill [66]

Answer:

Explanation:

The displacement is zero since it goes in a full circle and ends up where it started.

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2 years ago

Why does Quito, Equator has very little changes to the daylight hours<br> throughout the year?

marin [14]

Because of the position on the equator, the change in rotation of the Earth on its axis throughout the year doesn't affect it much. Unlike the poles, Quito is almost constantly in direct view of the sun. So, because of lack of change in rotation, the daylight hours are hardly varied as Quito is almost constantly in more or less the same spot in relation to the sun.

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3 years ago

Mary has a mass of 40 kg and sprints at 1 m/s. How much kinetic energy does she have?

Sav [38]

Answer:

20 J

Explanation:

Kinetic energy is given as half of the product of mass and the square of velocity of an object:

KE = $\frac{1}{2}mv^2$

where m = mass = 40 kg

v = velocity = 1 m/s

Hence, Mary's kinetic energy is:

KE = $\frac{1}{2} * 40 * 1^2$

KE = 20 * 1 = 20 J

She has a kinetic energy of 20 J.

8 0

3 years ago

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$