Answer:
CO2 molecule is made up of Carbon (atomic mass =12) and oxygen (atomic mass=16).
So first finding the mass of 1 molecule of CO2 which is equals to
= mass of 1 carbon atom + masses of 2 oxygen atom, we get
= 12+(16*2)= 12+32= 44 a.m.u.
Now 1 molecule of CO2 has mass 44 amu so mass of 1 mole CO2 will be 44 grams.( 1 a.m.u.=1.6729*10^-33 grams. 1 mole = 6.022*10^23, so 44 a.m.u.=73.6076*10^-33 grams approx. For one mole CO2, 73.6076*10^-33*6.022*10^23 which is approximately equals to 44 grams. )
1 mole CO2= 44grams, so 2.5 moles = 44*2.5= 110 grams
So our answer is 110 grams
The rate of doing work is called POWER.
Power is also defined as the rate of transfer of energy. It measures the speed of how work is done.
The unit of power is WATT.
Power is computed by dividing WORK DONE by TIME TAKEN.
P = W/t
1 watt:
*power of a machine
*power of an agent that does work at the rate of 1 joule per secon
1000 watt = 1 kilowatt
1000 kilowatt = 1 megawatt
Answer:
a) dB / dA = 2
,
b) Network B Network A
2 1
4 2
6 3
Explanation:
a) The expression for grating diffraction is
d sin θ = m λ
where d the distance between two slits, λ the wavelength and m an integer that represents the diffraction range
In this exercise we are told that the two spectra are in the same position, let's write the expression for each network
Network A
m = 1
sin θ = 1 λ / dA
Network B
m = 2
sin θ = 2 λ / dB
they ask us for the relationship between the distances, we match the equations
λ / dA) = 2 λ / dB
dB / dA = 2
b) let's write the equation of the networks
sin θ = m_A λ / dA
sin θ = m_B λ / dB
we equalize
m_A λ/ dA = m_B λ / dB
we use that
dB / dA = 2
m_A 2 = m_B
therefore the overlapping orders are
Network B Network A
2 1
4 2
6 3
The correct answer is A I believe but I may not be correct. Either A or B
Answer: The angular speed w of the loop = 181 rad/s
Explanation:
Given that;
Area A = 0.193 m2
Magnetic field B = 0.374 T
E.m.f = 9.24v
Ø = 45 degree
According to Faraday's law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.
Using the formula
E.m.f = NABwcosØ
Where w = angular velocity.
Let assume that N = 1 then,
9.24 = 0.193 × 0.374 × w × cos45
9.24 = 0.051w
w = 9.24/0.051
w = 181 m/s
Therefore, the angular speed w of the loop = 181 rad/s