The magnitude of a vertical electric field that will balance the weight of a plastic sphere of mass 2. 1 g that has been charged to -3. 0 NC is 6.86 × 
N/C.
An electric field is the physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field of a system of charged particles.
It is given that,
Mass of sphere, m = 2.1 g =0.0021kg
Charge,q = ₋3nC = ₋ 3 ₓ 
To balance the weight of a plastic sphere, we must determine the magnitude of the electric field. So,
a = g
E = 
E = 6860000 N/C
E = 6.86 × N/C
Hence, the magnitude of the electric field that balances its weight is 6.86 × N/C .
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Answer:
During a lunar eclipse, the Moon turns red because the only sunlight reaching the Moon passes through Earth's atmosphere. The more dust or clouds in Earth's atmosphere during the eclipse, the redder the Moon will appear.
Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
Explanation:
1)
A) Bb BB
B) 50%
2)
A) 50%
B) <u> </u><u> </u><u> </u><u> </u><u> </u><u>b</u><u>.</u><u> </u><u> </u><u> </u><u>b</u>
B. Bb. Bb
b. bb. bb
