Answer the assignment questions using the World Geology map in this lesson and/or the earth's lithospheric plates map in the ass
1 answer:
You might be interested in
Answer:
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
Explanation:
For this exercise we can use Gauss's law
Ф = ∫ E. dA = / ε₀
Where q is the charge inside the surface.
In this case the surface must be a sphere, the electric field lines and the radius of the sphere are parallel, so the scalar product is reduced to the algebraic product
∫ E dA = q_{int}/ ε₀
The area of a sphere is
A = 4π r²
- The electric field for a distance r < R_int
The charge inside is zero, so the electric field
E = 0 r <R_in t
- The field for a radio inside the shell
Let's use the concept of density
ρ = Q / V
q = ρ (4/3 π r³)
dq = ρ 4π r² dr
We substitute in the Gaussian equation
E ∫ dA = ρ 4π r² dr / ε₀
E 4π r² = ρ 4π/ε₀ r³ / 3
E = ρ / 3ε₀ r
We evaluate between the lower limit r = R_int, E = 0 and the upper limit r = r, E = E
E- 0 = ρ / 3ε₀ (r –R_int)
Density is
ρ = q / 4/3 π (R_out³ - R_int³)
Where R <r
E = Q / 4 π ε₀ (r-R_int) / (R_put³ -R_int³)
Answer:
a)
Nm⁻¹
b)
Hz
c)
ms⁻¹
d)
m
e)
ms⁻²
f)
m
Explanation:
a)
= force required to hold the object at rest connected with stretched spring = 27 N
= stretch in the spring from equilibrium position = 0.2 m
= force constant of the spring
force required to hold the object at rest is same as the spring force , hence
inserting the values
Nm⁻¹
b)
frequency of the oscillations is given as
inserting the values
Hz
c)
= Amplitude of oscillations = 0.2 m
= angular frequency
Angular frequency is given as
rads⁻¹
Maximum speed of oscillation is given as
ms⁻¹
d)
maximum speed of the object occurs at the equilibrium position, hence
m
e)
Maximum acceleration of oscillation is given as
ms⁻²
f)
maximum acceleration occurs when the object is at extreme positions, hence
m
The answer is letter C.
Weight (on Earth) is the force due to the mass of Earth attracting whatever mass is subject of discussion.
The force of attraction between any two masses is called Newton's Law of Universal Gravitation:
is simply a given constant.
If we're at the surface of Eath,
refers to the mass of the Earth, 
to the mass of whatever is on the surface of Earth, and 
to the radius of Earth.
Normally, we define a constant
to be equal to 
; in which 
is the mass of Earth and 
the radius of earth; happens to be around 9.8.
By that, we adapt the Law of Universal Gravitation to objects on the surface of Earth, we call that force Weight.
As you can see, weight is directly proportional to mass, more mass implies more weight.
Weight (on Earth) is the force due to the mass of Earth attracting whatever mass is subject of discussion.
The force of attraction between any two masses is called Newton's Law of Universal Gravitation:
If we're at the surface of Eath,
Normally, we define a constant
By that, we adapt the Law of Universal Gravitation to objects on the surface of Earth, we call that force Weight.
As you can see, weight is directly proportional to mass, more mass implies more weight.
Answer:
0.629 m
Explanation:
For destructive interference we have the case
m = 1,2,3.....
Frequency is given by
Wavelength
The angle
The width is
The width of the hole is 0.629 m
For destructive interference
The angles are