Answer:
The distance from the charge is 3.35 m.
Explanation:
Given that,
Electric potential, V = 635 V
Magnitude of electric field, E = 189 N/C
We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :
d is the distance from charge
So, the distance from the charge is 3.35 m. Hence, this is the required solution.
28.1 grams is 0.0281 kg.
43.278 kg - 0.0281 kg = 43.2499 or if you round -> 43.250
Answer:
The velocity of the astronaut is, v = - 0.36 m/s
Explanation:
Given data,
The mass of the astronaut, M = 84 kg
The mass of the gas expelled by the thruster, m = 35 g
= 0.035 g
The velocity of the gas expelled, v = 875 m/s
According to the conservation of momentum,
MV + mv = 0
V = - mv / M
Substituting the values,
V = - 0.035 x 875 / 84
= - 0.36 m/s
The negative sign in the velocity indicates the astronaut moves opposite to the direction of velocity of gas.
Hence, the velocity of the astronaut is, v = - 0.36 m/s
B. Primary target market , because its a business ( you sell in a business ) and it says a certain group which would be a target.
refraction is the right answer
