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wlad13 [49]
3 years ago
7

For elliptical obits: the direction of the velocity of the satellite is _______________________ (always, seldom, never) perpendi

cular to the net force acting upon the satellite. For circular obits: the direction of the velocity of the satellite is ___________________ (always, seldom, never) perpendicular to the net force acting upon the satellite.
Physics
1 answer:
alexandr1967 [171]3 years ago
7 0

Answer:

For elliptical orbits: seldom

For circular orbits: always

Explanation:

We start by analzying a circular orbit.

For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.

Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.

So for a circular orbit,

the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.

Now we analyze an elliptical orbit.

An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.

Therefore, for an elliptical orbit,

the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.

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A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex
irina1246 [14]

a)

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

F_{B_x}=0

F_{B_y}=0

b)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=0

F_{B_y}=3.21\cdot 10^{-3}N (+z axis)

c)

F_{E_x}=1.19\cdot 10^{-3} N (+x axis)

F_{B_x}=3.21\cdot 10^{-3} N (+y axis)

F_{B_y}=3.21\cdot 10^{-3}N (-x axis)

Explanation:

a)

The electric force exerted on a charged particle is given by

F=qE

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

q=+4.9\mu C=+4.9\cdot 10^{-6}C is the charge

E_x=+242 N/C is the electric field, along the x-direction

So the electric force (along the x-direction) is:

F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N

towards positive x-direction.

The magnetic force instead is given by

F=qvB sin \theta

where

q is the charge

v is the velocity of the charge

B is the magnetic field

\theta is the angle between the directions of v and B

Here the charge is stationary: this means v=0, therefore the magnetic force due to each component of the magnetic field is zero.

b)

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

F_{E_x}=1.19\cdot 10^{-3} N (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- B_x: this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=0^{\circ}, so the force due to this field is zero.

- B_y: this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, \theta=90^{\circ}. Therefore, \theta=90^{\circ}, so the force due to this field is:

F_{B_y}=qvB_y

where:

q=+4.9\cdot 10^{-6}C is the charge

v=345 m/s is the velocity

B_y = +1.9 T is the magnetic field

Substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

c)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

F_{E_x}=1.19\cdot 10^{-3}N (+x axis)

For the field B_x, the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

F_{B_x}=qvB_x

And by substituting,

F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field B_y, the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

F_{B_y}=qvB_y

And by substituting,

F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0
3 years ago
Hey guys.. ans me
Stels [109]

The time taken by the swimmer was 1 hour.

Why?

Since the swimmer is maintaining an angle of 150° while he was swimming, there were two components of the speed (horizontal and vertical). If we want to calculate the time taken by him to cross the river, we need to calculate the vertical speed and consider that the flow's speed is compensated by his horizontal speed.

We can calculate both components of the speed using the following formula:

HorizontalSpeed=Speed*Cos(150\°)\\\\VerticalSpeed=Speed*Sin(150\°)

Now, calculating we have:

HorizontalSpeed=2\frac{Km}{h} *Cos(150\°)=-\sqrt{3} \frac{Km}{h} \\\\VerticalSpeed=2\frac{Km}{h} *Sin(150\°)=1\frac{Km}{h}

Therefore, we have that the horizontal speed is compesating the flow's speed while his vertical speed is used to cross the river which is 1 Km wide.

Hence, we have that the tame taken is:

Time=\frac{RiverWidth}{VerticalSpeed}=\frac{1Km}{1\frac{Km}{h} } =1h

Have a nice day!

7 0
3 years ago
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