Find Displacement and Distance
displacement ...
north is 700+400+100 =1200m n
south=1200m
1200-1200=0
east is 300+300=600m
west is 600m
600-600=0
back at dtart. displ zero
distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m = 3600m
Answer and Explanation:
curents i = 2.9 A
i ' = 4.4 A
the magnitude (in T.m) of the path integral of B.dl around the window frame = μo * current enclosed
= μo* ( i '- i )
Since from Ampere's law
where μ o = permeability of free space = 4π * 10 ^-7 H / m
plug the values we get the magnitude (in T.m) of the path integral of B.dl = ( 4π*10^-7 ) (2.9+4.4)
= 1.884 * 10^-6 Tm
Answer:
a) 14.2 atm
b) 4.46 atm
c) 1.06 atm
Explanation:
For an ideal gas,
PV = nRT
P = pressure of the gas
V = volume occupied by the gas
n = number of moles of the gas
R = molar gas constant = 0.08206 L.atm/mol.K
T = temperature of the gas in Kelvin
a) For HF,
P =?, V = 2.5L, n = 1.35 moles, T = 320K
P = 1.35 × 0.08206 × 320/2.5
P = 14.2 atm
b) For NO₂
P =?, V = 4.75L, n = 0.86 moles, T = 300K
P = 0.86 × 0.08206 × 300/4.75
P = 4.46 atm
c) For CO₂
P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K
P = 2.15 × 0.08206 × 330/55
P = 1.06 atm
Answer:3.87*10^-4
Explanation:
What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca
We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.
C=f*lambda
3*10^8=f*3.44*10^-12
F=0.87*10^20 hz
Then with the frequency, find the energy emitted using equation
E=hf E = freq*Plank's constant
E=.87*10^20*6.62*10^-34
E=575.94*10^(-16)
With this energy, convert into MeV from joules.
With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.
Plugging and computing all necessary numbers gives you
3.87*10^-4 u.
