If earth began to shrink, but its mass remained the same, what would happen to the value of g on earth's surface?
1 answer:
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Answer:
The horizontal distance is 4.823 m
Solution:
As per the question:
Mass of man, m = 65.0 kg
Height of the hill, H = 5.00 m
Mass of the backpack, m' = 20.0 kg
Height of ledge, h = 2 m
Now,
To calculate the horizontal distance from the edge of the ledge:
Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.
Now,
where
KE = Kinetic energy
PE = Potential energy
Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.
Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.
Therefore,
where
v = velocity at the hill's bottom
Now,
Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:
Now, time taken for the fall:
Now, the horizontal distance is given by:
x = v't =
The electrical force acting on a charge q immersed in an electric field is equal to
where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
where
q is the charge
E is the strength of the electric field
In our problem, the charge is q=2 C, and the force experienced by it is
F=60 N
so we can re-arrange the previous formula to find the intensity of the electric field at the point where the charge is located:
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