The answer is Sagittarius A, a very large black hole.
Answer: Recall that
Work (W) = - e × ∆V
Where
Charge of electron= 1.6 × 10^ -19 c
And from the question we were given change in potential as = -0.0677V
Work = - (1.6 × 10^-19) × (-0.0677)
Work= 1.0832 × 10^-20J
Therefore the work done by the electric force when a sodium ion (charge = +e) moves from the outside to the inside is 1.0832×10^-20
Circumference : 4/10=0.4m
Since diameter = circumference / pi (3.14)
d = 0.4/pi = 0.127388m
So your answer is 0.127388m
Answer:
34.8 and 55.2º
Explanation:
This is a projectile launching exercise, as we are told that the range of the arrow must be equal to its range and = 31 m let's use the equation
The scope equation is
R = v₀² sin 2θ /g
sin 2 θ = R g / v₀²
sin 2 θ = 31 9.8 / 18²
2 θ = sin⁻¹ 0.93765
θ = 34.8º
At the launch of projectiles we have two complementary angles with the same range in this case 34.8 and (90-34.8) = 55.2º