T is in seconds (s)
<span>2pi is dimensionless </span>
<span>L is in meters (m) </span>
<span>g is in meters per second squared (m/s^2) </span>
<span>so you can write the equation for the period of the simple pendulum in its units... </span>
<span>s=sqrt(m/(m/s^2)) </span>
<span>simplify</span>
<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>
<span>s=sqrt(s^2) </span>
<span>s=s </span>
<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
Answer:41.991ml
Explanation:
Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION
2 H+ + 2e → H2 REDUCTION
Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.
from the oxidation reaction:
1moles of oxygen requires 4moles of electrons to be discharged at the product
F=96500C/mol
Quantity of charge Q=It
=60*60*0.201A
Q=723.6C
Mole=Q/(F*mole ratio of electron)
Mole= 723.6/(4*96500)
Mole=((1809)/(965000))
M=0.0018746114
M1/M2=V1/V2
1/0.00187=22.4dm^3/V2
V2=22.4*0.00187
V2=0.04199129534dm^3
41.99129534ml
Answer:
frequency is 195.467 Hz
Explanation:
given data
length L = 4.36 m
mass m = 222 g = 0.222 kg
tension T = 60 N
amplitude A = 6.43 mm = 6.43 × 
m
power P = 54 W
to find out
frequency f
solution
first we find here density of string that is
density ( μ )= m/L ................1
μ = 0.222 / 4.36
density μ is 0.050 kg/m
and speed of travelling wave
speed v = √(T/μ) ...............2
speed v = √(60/0.050)
speed v = 34.64 m/s
and we find wavelength by power that is
power = μ×A²×ω²×v / 2 ....................3
here ω is wavelength put value
54 = ( 0.050 ×(6.43 × )²×ω²× 34.64 ) / 2
0.050 ×(6.43 × )²×ω²× 34.64 = 108
ω² = 108 / 7.160 × 
ω = 1228.16 rad/s
so frequency will be
frequency = ω / 2π
frequency = 1228.16 / 2π
frequency is 195.467 Hz
Emissivityis a measure of how much thermal radiation a body emits to its environment. On the other hand we have that reflectivity is a measure of how much is reflected, and transmissivity is a measure of how much passes through the object. If a body is required to be ideally reflective to its maximum efficiency, the body should NOT have the property of transmissivity or emissivity. Therefore it should be 0 its emittivity.
Correct answer would be A : ZERO.
