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shtirl [24]
2 years ago
5

The caste system is an example of​

Physics
1 answer:
snow_lady [41]2 years ago
7 0

Answer:

It is example of Closed system

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A point charge is placed 100 m from a 6 uC charge generating an electric field. What is the
kvv77 [185]

The strength of the electric field is 5 N/C

Explanation:

The magnitude of the electric field produced by a single-point charge is given by:

E=\frac{kQ}{r^2}

where

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

Q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

Q=6 \mu C = 6\cdot 10^{-6}C is the charge producing the field

r = 100 m is the distance from the charge at  which we want to calculate the field

Substituting into the equation, we find the s trength of the electric field:

E=\frac{(8.99\cdot 10^9)(6\cdot 10^{-6})}{(100)^2}=5.4 N/C \sim 5 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0
3 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
Why it is difficult to run fast in sand​
balu736 [363]
Running on sand requires 1.6 times more energy spent than running on hard surface, so the force applied by our foot on sand is less.
8 0
3 years ago
Read 2 more answers
The energy emitted from the sun is a product of ________.
Tamiku [17]
Q. The energy emitted from the sun is a product of ________.

A. Fusion
3 0
3 years ago
Natalie accelerate her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 s. find her average acceleration
Vikentia [17]
Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s

acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²

Answer: the acceleration would be 1.6 m/s²
8 0
3 years ago
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