Answer:
total work is 99.138 kJ
Explanation:
given data
diameter = 5 cm
depth = 75 m
density = 1830 kg/m³
to find out
the total work
solution
we know mass of volume is
volume = 
volume = 
so
work required to rise the mass to the height of x m
dw = 
gx dx
so total work is integrate it with 0 to 75
w = 
w = 
× 0.05² × 1830× 9.81× 
w = 99138.53 J
so total work is 99.138 kJ
Explanation:
work done by friction = 1/2 x 42 x ( 3.33^2 - 11.5^2)
= 21 ( 11.08 - 132.25)
= 21 ( - 121.17 )
= - 2544.57 J
Complete question:
A uniform electric field is created by two parallel plates separated by a
distance of 0.04 m. What is the magnitude of the electric field established
between the plates if the potential of the first plate is +40V and the second
one is -40V?
Answer:
The magnitude of the electric field established between the plates is 2,000 V/m
Explanation:
Given;
distance between two parallel plates, d = 0.04 m
potential between first and second plate, = +40V and -40V respectively
The magnitude of the electric field established between the plates is calculated as;
E = ΔV / d
where;
ΔV is change in potential between two parallel plates;
d is the distance between the plates
ΔV = V₁ -V₂
ΔV = 40 - (-40)
ΔV = 40 + 40
ΔV = 80 V
E = ΔV / d
E = 80 / 0.04
E = 2,000 V/m
Therefore, the magnitude of the electric field established between the plates is 2,000 V/m
<span>since sound travels using mechanical waves and needs a material medium to propagate and since mechanical waves spread through vibrations ...and since hard materials have their atoms packed closely....they need to vibrate with a smaller amplitude to pass on the wave....thus sound travels faster in a denser medium than a less dense one.</span>
