Answer:
Tips for Balancing Chemistry Equations You may only put numbers in front of the molecules. You can never alter the subscript, as it would alter the formula. ... Begin by counting and making note of the elements present on each side of the equation. Count the number of molecules of each element on both sides of the equation. More items...
Explanation:
I dont know what the question is
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
Answer:
Explanation:
<u>Charge of an Electron</u>
Since Robert Millikan determined the charge of a single electron is
Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is 
Let's test the possible charges listed in the question:
. We have just found it's a possible charge of a particle
. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets
this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge
cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6
Finally, the charge 
is four times the charge of the electron, so it is a possible value for the charge of an oil droplet
Summarizing, the following are the possible values for the charge of an oil droplet:
Answer:
(a) 1462.38 m/s
(b) 2068.13 m/s
Explanation:
(a)
The Kinetic energy of the atom can be given as:
K.E = (3/2)KT
where,
K = Boltzman's Constant = 1.38 x 10⁻²³ J/k
K.E = Kinetic Energy of atoms = 343 K
T = absolute temperature of atoms
The K.E is also given as:
K.E = (1/2)mv²
Comparing both equations:
(1/2)mv² = (3/2)KT
v² = 3KT/m
v = √[3KT/m]
where,
m = mass of Helium = (4 A.M.U)(1.66 X 10⁻²⁷ kg/ A.M.U) = 6.64 x 10⁻²⁷ kg
v = RMS Speed of Helium Atoms = ?
Therefore,
v = √[(3)(1.38 x 10⁻²³ J/K)(343 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 1462.38 m/s</u>
(b)
For double temperature:
T = 2 x 343 K = 686 K
all other data remains same:
v = √[(3)(1.38 x 10⁻²³ J/K)(686 K)/(6.64 x 10⁻²⁷ kg)]
<u>v = 2068.13 m/s</u>
