If an astronaut throws an object in space, the object’s speed will _____
1 answer:
The object's speed will not change.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.
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Part A)
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
And therefore, the energy density at this distance from the center is
First of all, let's convert the radii of the inner and the outer sphere:
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
from which
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
And so, the energy density at r=0.111 m is
Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:
And therefore, the energy density at this distance from the center is