Anybody please help me with these questions... I really need help...
1 answer:
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Answer:
P = 1471500 [Pa]
Explanation:
We must remember that pressure is defined as the relationship between Force over the area.
where:
P = pressure [Pa] (units of pascals)
F = force [N] (units of Newtons)
A = area of contact = 4 [cm²]
But first we must convert from cm² to m²
A = 0.0004 [m²]
Also, the weight should be calculated as follows:
where:
m = mass = 60 [kg]
g = gravity acceleration = 9.81 [m/s²]
Now replacing:
And the pressure:
Because 1 [Pa] = 1 [N/m²]
2000J
Explanation:
Given parameters:
Extension = 0.5m
Spring constant = 16000N/m
Unknown:
Energy stored in the bow string = ?
Solution:
The energy stored in a bow string is an elastic potential energy.
It can be calculated using the expression below;
Elastic energy = K e²
Where k is the spring constant
e is the extension
Input the parameters;
Elastic energy = K e²
= x 16000 x 0.5²
= 2000J
learn more:
Potential energy brainly.com/question/10770261
#learnwithBrainly
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