The sun is the center of the solar system which is surrounded by the nine planets and they tend to orbit the sun in concentric circles with the sun as the center.
<h3>What is the solar system?</h3>
The term solar system has to do with the arrangement of the sun and the planets. The sun lies at the focus of the solar system. Now we know that there are nine planets that orbit around the sun. The distance between the planets and the sun depends on their relative proximity to each other.
Thus, the sun is the center of the solar system which is surrounded by the nine planets and they tend to orbit the sun in concentric circles with the sun as the center.
Learn more about solar system:brainly.com/question/12075871
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Answer:
F = 352 N
Explanation:
we know that:
F*t = ΔP
so:
F*t = M
-M
where F is the force excerted by the wall, t is the time, M the mass of the ball, the final velocity of the ball and the initial velocity.
Replacing values, we get:
F(0.05s) = (0.8 kg)(11m/s)-(0.8 kg)(-11m/s)
solving for F:
F = 352 N
Answer:
I think like 2024 or 25
Explanation:
Elon musk will probably go
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.
