Answer:
E = 1,873 10³ N / C
Explanation:
For this exercise we can use Gauss's law
Ф = E. dA =
/ ε₀
Where q_{int} is the charge inside an artificial surface that surrounds the charged body, in this case with the body it has a spherical shape, the Gaussian surface is a wait with radius r = 1.35 m that is greater than the radius of the sphere.
The field lines of the sphere are parallel to the radii of the Gaussian surface so the scald product is reduced to the algebraic product.
The surface of a sphere is
A = 4π r²
E 4π r² = q_{int} /ε₀
The net charge within the Gauussian surface is the charge in the sphere of q1 = + 530 10⁻⁹ C and the point charge in the center q2 = -200 10⁻⁹ C, since all the charge can be considered in the center the net charge is
q_{int} = q₁ + q₂
q_{int} = (530 - 200) 10⁻⁹
q_{int} = 330 10⁻⁹ C
The electric field is
E = 1 / 4πε₀ q_{int} / r²
k = 1 / 4πε₀
E = k q_{int}/ r²
Let's calculate
E = 8.99 10⁹ 330 10⁻⁹/ 1.32²
E = 1,873 10³ N / C
Answer:
Stop cheating in exam
Explanation:
Shame!!!!
I am sorry but I will have to refer you to the student conduct at UTA.
Answer:
none
Explanation:
~both of them show to the nearest metre.
~millimeter has (mm) unit eg 0.7mm
Answer:
Technician A
Explanation:
camshafts rotate at one-half the crankshaft speed
Answer:
What is the best description for the volume of air volume of air provided in a high quality rescue breath?
Explanation:
A) only Enough air to create a visible rise of the chest.
B) Until you can no longer force air in.
C) plenty of air make sure it is adequate to sustain life
D) clear and obvious rise of the chest, sustained over a few seconds