Answer:
a) 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b) 3.466 × 10¹¹ N/C
Explanation:
a)
p(r) = -A exp ( - 2r/a₀)
Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV = -A ₀∫^∞ ₀∫^π ₀∫^2π exp ( - 2r/a₀)r² sinθdrdθd∅
Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e
now using integration by parts;
A = e / πa₀³
p(r) = - (e / πa₀³) exp (-2r/a₀)
Now Net charge inside a sphere of radius a₀ i.e Qnet is;
= e - (e / πa₀³) ₀∫^a₀ ₀∫^π ₀∫^2π r² exp (-2r/a₀)dr
= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C ( e = 1.6 × 10⁻¹⁹C)
b)
Using Gauss's law,
E × 4πa₀ ² = Qnet / ∈₀
E = 4πa₀ ² × Qnet × 1/a₀²
E = 3.466 × 10¹¹ N/C
Answer:
The unit you should use for work done and energy is the joule (J) which is indeed the same as the newton metre (N m).
There is another physical quantity which is the product of force and distance and that is torque or moment of a force.
The unit you should use for torque is the newton metre (Nm) and not the joule.
Naming the units of work done and torque differently helps to emphasis the fact that work done and torque refer to two different physical quantities although the definitions of both quantities have the product of force and distance in them.
work done=force→⋅displacement→ and torque→=force→×displacement→
Hope I helped
Here we can use the work energy theorem
here we know that
as it come to rest finally
now work done by friction force will be given as
Work done by spring force is given as
so now plug in all data above
so above is the friction coefficient
Answer:
The correct option is D
Explanation:
In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.
They'll vibrate at their characteristic resonant frequency. That depends on the material the object is made of and its shape.
