The magnitude of the electric field at a point inside the sphere that lies 8.0 cm from the center is 4.5 ×
Given
Radius = 20 cm
Electric field at a point inside the sphere that lies 8.0 cm from the center.
ρ = 1.5 x 10-6 c/m3
E (r) = ρr / 3εο = 1.5 × × 8 × / 3 × 8.85 × = 4.5 ×
The physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also referred to as an E-field[1]). [2] It also describes the physical field of a system of charged particles. [3] Electric charges and time-varying electric currents are the basis of electric fields. A manifestation of the electromagnetic field, one of the four fundamental interactions (sometimes known as forces) of nature, is electromagnetic fields, which also include electric and magnetic fields.
Electric fields play a significant role in many branches of physics and are used in electrical engineering. For instance, in atomic physics and chemistry, the electric field acts as the attracting force that holds the atomic nucleus and electrons together in atoms. Additionally, it is the driving force for atoms' chemical bonds.
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Answer:
I found the experience tasking
Explanation:
I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.
I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn
First let's find the time it takes for the first ball to land:
Acceleration is a=-g so vertical velocity is V=-gt + V1sin(30).
Position is thus
S=(-1/2)gt^2 +V1t sin(30).
Solving for t gives
t=2V1sin(30)/g
The second ball has the same position function except for the new velocity, which is given by
V2=2V1. Putting this in and solving for t2 gives
t2=4V1sin(30)/g.
It takes twice as long for the second ball to land on the ground.
The horizontal distance of ball 1 is S1 = V1t cos(30). Again we look at ball 2's distance by substituting V2=2V1 and get
S2 = 2V1t2 cos(30).
Note here I put in t2 since it will fly for that amount of time. But we already saw that
t2 = 2t1
So S2=4V1 cos(30)
That is the second ball goes 4 times further than the first one. This is because it is going twice as fast along both the horizontal and the vertical. It moves horizontally twice as fast for twice as long.
14 years (;
1 light year is how long it takes for a light wave to travel 1 year
I would guess anything that is black? black absorbs there most light energy. that's is why a black car in the sun will always be hotter inside than a white car.