Why is Electricity an important part of our daily life ???
2 answers:
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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m
Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:
Substitute numerical values:
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
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Answer:
a) a = 6.1 m/s^2
b) a = 0.98m/s^2
Explanation:
Mass of slab = 40kg
Mass of block = 10kg
Coefficient of static friction (Us) = 0.60
Kinetic coefficient (UK) = 0.40
Horizontal force = 100N
The normal reaction from 40kg slab on 10 kg block = 10*9.81
= 98.1N
Static frictional force = Us*R
= 98.1*0.6
= 58.86N
This is less than the force applied
If 10 kg block will slide on the 40 kg slab, net force = 100 - kinetic force
Kinetic force (Uk*R) = 0.4*98.1
= 39.28N
= 39N
Net force = 100 -39
= 61N
Recall that F = ma
For 10 kg block
a = F/m
a = 61/10
a = 6.1m/s^2
b) Frictional force on 40 kg slab by 10 kg = 98.1*0.4
= 39.24
= 39N
F = ma
a = F/m
For 40kg slab
a = 39/40
a = 0.98m/s^2
