Answer:
E = 12640.78 N/C
Explanation:
In order to calculate the electric field you can use the Gaussian theorem.
Thus, you have:

ФE: electric flux trough the Gaussian surface
Q: net charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m
Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

Finally, you obtain for E:

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C
Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
Answer:
<em>➢</em><em>when you crank you make kinetic energy and then the kinetic energy makes potential energy.</em>
Explanation:
<em>hope</em><em> it</em><em> will</em><em> help</em><em> you</em><em> have</em><em> a</em><em> great</em><em> day</em><em> bye</em><em> and</em><em> Mark</em><em> brainlist</em><em> if</em><em> the</em><em> answer</em><em> is</em><em> correct</em>
<em>
</em>
<em>#</em><em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em> </em><em>on </em><em>learning</em>
12.00 min = 0.2 hr
8.00 min = 0.15 hr
Total distance:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)
= 8.25 km
Average speed:
(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3
= 15 km/hr
Change in position:
(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)
= 0.25 km
Average velocity:
(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3
≈ 1.67 m/s
Answer:
the results will be the same.it may be