<span>Nuclear energy can be used to power all of
the above choices. Nuclear power plants produce radioactive waste that must be
stored properly. It is very impossible for a nuclear power plant to have no
waste at all since lots of chemicals are used to create the process as it gives
energy to other machines, weapons such as bombs and powering submarines. Radioactive waste can not be released into
local water supplies since the wastes are very radioactive and may cause
mutation to the fishes and bioaccumulation which will affect humans as well. It
will also cause air pollution if the chemicals are not stored properly.</span>
Answer:
(
)=1913.31 N/m^2
Explanation:
given:
=0.85
=90 m/s
γ∞=1.23 kg/m^3
solution:
since outside pressure is atm pressure vaccum can be defined by (
)
=√2(
)/γ∞[
-1]
(
)=1913.31 N/m^2
Answer:
induced EMF = 240 V
and by the lenz's law direction of induced EMF is opposite to the applied EMF
Explanation:
given data
inductance = 8 mH
resistance = 5 Ω
current = 4.0 A
time t = 0
current grow = 4.0 A to 10.0 A
to find out
value and the direction of the induced EMF
solution
we get here induced EMF of induction is express as
E = - L
...................1
so E = - L 
put here value we get
E = - 8 ×

E = -40 × 6
E = -240
take magnitude
induced EMF = 240 V
and by the lenz's law we get direction of induced EMF is opposite to the applied EMF
Answer:
Explanation:
a )
Reaction force of the ground
R = mg
= 160 N
Maximum friction force possible
= μ x R
= μ x 160
= .4 x 160
= 64 N .
b )
160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,
Taking moment about top point of ladder
160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3
240 + 444 + 4f = 2700
f = 504 N
c )
Let x be the required distance.
Taking moment about top point of ladder
160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )
240 + 444 x + 1440 = 2700
x = 2.3 m
so man can go upto 2.3 at which maximum friction acts .
Answer:
0.34 m
Explanation:
From the question,
v = λf................ Equation 1
Where v = speed of sound, f = frequency, λ = Wave length
Make λ the subject of the equation
λ = v/f............... Equation 2
Given: v = 340 m/s, f = 500 Hz.
Substitute these values into equation 2
λ = 340/500
λ = 0.68 m
But, the distance between a point of rarefaction and the next compression point, in the resulting sound is half wave length
Therefore,
λ/2 = 0.68/2
λ/2 = 0.34 m
Hence, the distance between a point of rarefaction and the next compression point, in the resulting sound is 0.34 m