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ivanzaharov [21]
3 years ago
9

Which event occurred approximately 4.6 billion years ago

Physics
1 answer:
Levart [38]3 years ago
4 0
The answer to this is formation of earth and our solar system
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Assignment: Can you identify various forces and instances in which electrostatic and magnetic forces occur​
serg [7]

Answer:

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion. It is the basic force responsible for such effects as the action of electric motors and the attraction of magnets for iron. Electric forces exist among stationary electric charges; both electric and magnetic forces exist among moving electric charges. The magnetic force between two moving charges may be described as the effect exerted upon either charge by a magnetic field created by the other.

3 0
2 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
Pls helppp. Is this right?
andreyandreev [35.5K]

Answer:

yes you are totally right

7 0
3 years ago
Read 2 more answers
assuming birdman flies at height of 72m, how fast should he fly to hit bucket at 63m from start of field. gravity is -9.8m/s^2 n
algol [13]

Answer:

.....Birdman?

Explanation:

7 0
3 years ago
A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

W=K_f - K_i

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

K_f is the final kinetic energy

K_i = 66,120 J is the initial kinetic energy

Solving,

K_f = K_i + W = 66,120 + (-36,733)=29,387 J

Now we can find the final speed of the car by using the formula for kinetic energy

K_f = \frac{1}{2}mv^2

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s

3 0
3 years ago
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