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).
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A=mgh
m=300g=0.3kg
g=9,81 m/s^2
h=10m
A=29.43J
Th statement represented above : In a solution, the solvent is present in the larger amount is definitely FALSE. I choose this answer as a correct one as it also could be a solutions of alcohol ot for example <span>walter which means it's not so solid and liquid respectively. Hope it's clear! Regards!</span>
Answer:
The magnitude of the electric flux is 
Explanation:
Given that,
Electric field = 2.35 V/m
Angle = 25.0°
Area 
We need to calculate the flux
Using formula of the magnetic flux


Where,
A = area
E = electric field
Put the value into the formula



Hence, The magnitude of the electric flux is 
Answer:
f = 878,080 N
Explanation:
mass of pile driver (m) = 2100 kg
distance of pile driver to steel beam (s) = 5 m
depth of steel driven (d) = 12 cm = 0.12 m
acceleration due to gravity (g0 = 9.8 m/s^{2}
calculate the average force exerted on the pile driver by the beam.
- from work done = force x distance
- work done = change in potential energy of the pile driver
- equating the two equations above we have
force x distance = m x g x (s - d)
f x 0.12 = 2100 x 9.8 x (5- (-0.12))
d = - 0.12 because the steel beam went down at we are taking its
initial position to be an origin point which is 0
f = ( 2100 x 9.8 x (5- (-0.12)) ) ÷ 0.12
f = 878,080 N