The suns energy reaches the sun by Conduction Convection Radiation Fusion
2 answers:
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Answer:
a) -1.27*10³ N/C b) 0 c) -0.21*10³ N/C d) 0.1*10³ N/C
Explanation:
a) r = 0.76R
As this distance is inside the sphere, we need to know how much charge is enclosed within this distance for the center, as follows:
Q = ρ*V(r) = ρ*
where r = 0.760* R = 0.760* 0.295 m = 0.224 m, and ρ = -151 nC/m³
Applying Gauss' Law to a spherical gaussian surface of r= 0.76R, as the electric field is radial, and directed inward, we can write the following equation:
E*A = Q/ε₀, where Q= -7.11 nC, A= 4*π*(0.76R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = -1.27*10³ N/C
b) r= 3.90 R
As this distance falls inside the conducting shell, and no electric field can exist within a conductor in electrostatic condition, E=0
c) r = 2.8 R
As this distance falls between the sphere and the inner radius of the shell, we can calculate the electric field, applying Gauss' law to a gaussian surface of radius equal to r= 2.80 R.
First we need to find the total charge of the sphere, as follows:
Q = ρ*V =
In the same way that for a) we can write the following expression:
E*A = Q/ε₀, where Q= -16.2 nC, A= 4*π*(2.8R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = -0.21*10³ N/C
d) r= 7.30 R
In order to find the electric field at this distance, which falls beyond the outer radius of the shell, we need to find the total charge on the outer surface.
As the sphere has a charge of -16.2 nC, and the total charge of the conducting shell is 66.7nC, in order to make E=0 inside the shell, the total charge enclosed by a gaussian surface with a radius larger than the inner radius of the shell and shorter than the outer one, must be zero, which means that a charge of +16.2 nC must be distributed on the inner surface of the shell.
This leaves an excess charge on the outer surface of the shell as follows:
Qsh = 66.7 nC - 16.2 nC = 50.5 nC
Now, we can repeat the same process than for a) and c) as follows:
E*A = Q/ε₀, where Q= 50.5 nC, A= 4*π*(7.3R)² and ε₀ =8.85*10⁻¹² C²/N*m²
We can solve for E, as follows:
⇒ E = 0.1*10⁻³ N/C
Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ = L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Answer:
Magnitude of vector A = 0.904
Explanation:
Vector A , which is directed along an x axis, that is
Vector B , which has a magnitude of 5.5 m
The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A
Comparing we will get
Substituting in
So we have
Magnitude of vector A = 0.904