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kvv77 [185]
3 years ago
5

Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is

elliptical?
Physics
1 answer:
Leno4ka [110]3 years ago
6 0

Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.

This led to Kepler's law of orbital motion.

First Law: Planetary orbits are elliptical with the sun at a focus.

Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.

Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.

It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.

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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
What is the Unit used to measure voltage
muminat
The derived unit for voltage is named volt.
3 0
3 years ago
A styrofoam container used as a picnic cooler contains a block of ice at 0°C. If 325 g of ice melts in 1 hour, how much heat ene
Sauron [17]

Answer:

30.0625 W

Explanation:

325 g/h   x    (1h x 1kg)/(3600s x 1000g)   x   3,33 x 10^5 J/Kg = 30.0625 J/Kg = 30.0625 W

8 0
3 years ago
A coin is made of ______​
lys-0071 [83]

Answer:

coins made with the combination of copper zinc and Nickel

A cloth is made of fabrics

A towel is made from cotton

an umbrella is made up of fiberglass

a sweater is made up of wool

hope it helps

please mark me as the brainliest

5 0
3 years ago
If the charges attracting each other in the preceding problem have equal magnitudes,show that the magnitude of each charge is 2.
Schach [20]

Answer:

The magnitude of each charge is 2.82\times10^{-6}\ C

Explanation:

Suppose the two point charges are separated by 6 cm. The attractive force between them is 20 N.

We need to calculate the magnitude of each charge

Using formula of force

F=\dfrac{kq^2}{r^2}

Where, q = charge

r = separation

Put the value into the formula

20=\dfrac{9\times10^{9}\times q^2}{(6\times10^{-2})^2}

q^2=\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}

q=\sqrt{\dfrac{(6\times 10^{-2})^2\times20}{9\times10^{9}}}

q=2.82\times10^{-6}\ C

Hence, The magnitude of each charge is 2.82\times10^{-6}\ C

6 0
3 years ago
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