Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is
Explanation:
From the question we are told that
The radius of the inner circle is 
The radius of the outer circle is 
The charge on the spherical shell
The magnitude of the point charge at the center is 
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as

substituting values

where k is the coulomb constant with value 
substituting values


The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,

Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit,
: but as we saw from the previous equation, the orbital period of the Earth is proportional to
, therefore as
increases, T increases as well.
Therefore, the period of the orbit would increase.
Learn more about Kepler's third law:
brainly.com/question/11168300
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Answer:
= +3,394 103 m / s
Explanation:
We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.
The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s
The moment before the explosion
p₀ = (m₁ + m₂) v₀
After the explosion
pf = m₁
+ m₂ 
p₀ = [texpv_{f}[/tex]
(m₁ + m₂) v₀ = m₁
+ m₂
Let's calculate the final speed (v1f) of the first stage
= ((m₁ + m₂) v₀ - m₂
) / m₁
= ((2100 +1160) 4300 - 1160 5940) / 2100
= (14,018 10 6 - 6,890 106) / 2100
= 7,128 106/2100
= +3,394 103 m / s
come the same direction of the final stage, but more slowly