The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
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If the mass of the sun is 1x, at least one planet will fall into the habitable zone. if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.
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What are planets?</h3>
Planets are the large spherical shaped objects that rotate about the Sun in the elliptical orbits.
Planets are shaped from Planetary cloud. The dust storm and gases gathers under its own weight. The dense matter beginnings pivoting at high paces and accumulates more mass. The center structures, the star and rest of it ultimately levels into a curved plate from which planet is formed.
Thus, if I place a planet in orbits 1, 3, 5 , 6 and all planets will orbit the sun successfully.
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Answer:
Sound travels faster in liquids than in gases because molecules are packed more closely together. This means that when the water molecules begin to vibrate, they quickly begin to collide with each other forming a rapidly moving compression wave. Sound travels over four times faster than in air
Explanation:
Answer:

Explanation:
We are given that







We have to find the exit temperature.
By steady energy flow equation



Substitute the values




Answer:
0.6kg
Explanation:
the unknown here is the mass of the second block
applying the law of the conservation of momentum
m₁v₁ + m₂v₂ = (m₁ + m₂) v₃
where m₁=mass of first block=2.2kg
m₂=mass of colliding block= ?
v₁= velocity of first block=1.2m/s
v₂=velocity of colliding block=4.0m/s
v₃= final velocity of combined block=1.8m/s
applying the formula above
(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8
2.64 + 4m₂ = 3.96 + 1.8m₂
collecting like terms
4m₂ - 1.8m₂ = 3.96 - 2.64
2.2m₂=1.32
divide both sides by 2.2
m₂= 0.6kg