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:
The acceleration motorcycle
a = 5.13 m / s²
Explanation:
Now to determine the acceleration of the motorcycle
Use the force to analysis motion
∑ F = m * a
∑ F = E - D - m*g * sin ( β ) = m * a
E = 3168 N
D = 230 N
β = 31.6 °
3168 N - 230 N - 286 kg * 9.8 m / s² * sin ( 31.6° ) = 286 kg * a
Now solve to a'
a = [ 3168 N - 230 N - 286 kg * 9.8 m / s² * sin ( 31.6° ) ] / (286 kg)
a = 5.13 m / s²
Answer:
The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²
The rocket's motion for analysis sake is divided into two phases.
Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m
Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.
Explanation:
The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.
The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.
The detailed step by step solution to the problems can be found in the attachment below.
Thank you and I hope this solution is helpful to you. Good luck.
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The answer you are looking for is:
D.) It facilitates the moment of the current though a wire.
Hope that helps!!
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