Which element in the 18th column of the periodic table (shown below) has the largest radius?
1 answer:
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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:
<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>
Explanation:
<u>Physical Power </u>
It measures the amount of work W an object does in certain time t. The formula needed to compute power is
Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F
The second Newton's law gives us the net force as
being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:
Solving for a
The distance traveled in the interval is given by
Since vo=0
The force is given by
We don't know the value of m, so the force is
Computing the work done by the sprinter
The power is finally computed
During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration
The distance is
The net force is
The work done by the sprinter is now computed as
At last, the output power is
By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s