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Delvig [45]
3 years ago
13

How are important properties of Mercury, Venus, and Mars different from important properties of Earth?

Physics
1 answer:
yanalaym [24]3 years ago
3 0
Hi there,

There is one thing that is the most difference about other planets. Earth is a perfect condition for humans, animals, and other living things. Venus and Mars are not quite good for that. Venus has a very HOT surface, which makes it nearly impossible to be on. Mars on the other hand is very COLD, colder than earth because it is farther from the sun.
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The aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10−5t, collide with molecules of t
andreev551 [17]
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.

The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.

When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.

a)

Magentic force, F = q*v*B

q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T

Centripetal force, F = m*Ac = m * v^2 / R

where,

Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit

Now equal the two forces: q*v*B = m * v^2 / R => R =  m*v / (q*B)

=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]

=> R = 0.114 m

b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.

R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]

=> R = 10.4 m

 

4 0
3 years ago
Professor Martinez would like to describe the amount of variation there is in his data set. Which of the following should he cal
Finger [1]
Is it not standard deviation? or am i just dumb lol
7 0
3 years ago
Read 2 more answers
Mark creates a graphic organizer to review his notes about electrical force. Which labels belong in the regions marked X and Y?
sveta [45]

Answer:

The correct answer is A

Explanation:

The question requires as well the attached image, so please see that below.

Coulomb's Law.

The electrical force can be understood by remembering Coulomb's Law, that  describes the electrostatic force between two charged particles. If the particles have charges q_1 and q_2, are separated by a distance r and are at rest relative to each other, then its electrostatic force magnitude on particle 1 due particle 2 is given by:

|F|=k \cfrac{q_1 q_2}{r^2}

Thus if we decrease the distance by half we have

r_1 =\cfrac r2

So we get

|F|=k \cfrac{q_1 q_2}{r_1^2}

Replacing we get

|F|=k \cfrac{q_1 q_2}{(r/2)^2}\\|F|=k \cfrac{q_1 q_2}{r^2/4}

We can then multiply both numerator and denominator by 4 to get

|F|=k \cfrac{4q_1 q_2}{r^2}

So we have

|F|=4 \left(k \cfrac{q_1 q_2}{r^2}\right)

Thus if we decrease the distance by half we get four times the force.

Then we can replace the second condition

q_{2new} =2q_2

So we get

|F|=k \cfrac{q_1 q_{2new}}{r_1^2}

which give us

|F|=k \cfrac{q_1 2q_2}{r_1^2}\\|F|=2\left(k \cfrac{q_1 q_2}{r_1^2}\right)

Thus doubling one of the charges doubles the force.

So the answer is A.

8 0
3 years ago
Read 2 more answers
Help . . . . . . . . . . . . . . . . . . . .
Zigmanuir [339]

Explanation:

Transmission

Reflection

Absorption

Refraction

Diffraction

4 0
3 years ago
The diagram shows the field lines near the poles, X and Y, of two magnets.
Basile [38]

<em>Answer:  I think it is C</em>

Explanation: sorry if it wrong

3 0
3 years ago
Read 2 more answers
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