The formula that can be used to obtain a convergent series is given by;
a/1 – r.
<h3>What is a series?</h3>
In mathematics, we define a series as sum of numbers which could be convergent or divergent. In a convergent series, the summation of the numbers approaches a given value.
The formula that can be used to obtain a convergent series is given by;
a/1 – r.
Learn more about a convergent series:brainly.com/question/15415793?
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Answer:
The speed of the 270g cart after the collision is 0.68m/s
Explanation:
Mass of air track cart (m1) = 320g
Initial velocity (u1) = 1.25m/s
Mass of stationary cart (m2) = 270g
Velocity after collision (V) = m1u1/(m1+m2) = 320×1.25/(320+270) = 400/590 = 0.68m/s
Answer:
a) P=0.25x10^-7
b) R=B*N2*E
c) N=1.33x10^9 photons
Explanation:
a) the spontaneous emission rate is equal to:
1/tsp=1/3 ms
the stimulated emission rate is equal to:
pst=(N*C*o(v))/V
where
o(v)=((λ^2*A)/(8*π*u^2))g(v)
g(v)=2/(π*deltav)
o(v)=(λ^2)/(4*π*tp*deltav)
Replacing values:
o(v)=0.7^2/(4*π*3*50)=8.3x10^-19 cm^2
the probability is equal to:
P=(1000*3x10^10*8.3x10^-19)/(100)=0.25x10^-7
b) the rate of decay is equal to:
R=B*N2*E, where B is the Einstein´s coefficient and E is the energy system
c) the number of photons is equal to:
N=(1/tsp)*(V/C*o)
Replacing:
N=100/(3*3x10^10*8.3x10^-19)
N=1.33x10^9 photons
Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1
Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2
Now using formula of effective degree of freedom of mixture, (f), we have:
f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)
We know f=11/6,
substituting the value in above relation, we have:
Cv= R. 11/3*2
= R. 11/6
Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
= 17/6 R
Therefore, Cp/Cv = 17/11
Answer:Twice of given mass
Explanation:
Given
Two Particles of Equal mass placed at the base of an equilateral Triangle
let mass of two equal masses be m and third mass be m'
Taking one of the masses at origin
Therefore co-ordinates of first mass be (0,0)
Co-ordinates of other equal mass is (a,0)
if a is the length of triangle
co-ordinates of final mass 
Given its center of mass is at midway between base and third vertex therefore
