Rutherford's experiment<span> utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that </span>atoms<span> had an inner core which contained most of the mass of an </span>atom<span> and was positively charged.</span>
Answer:
∆T = Mv^2Y/2Cp
Explanation:
Formula for Kinetic energy of the vessel = 1/2mv^2
Increase in internal energy Δu = nCVΔT
where n is the number of moles of the gas in vessel.
When the vessel is to stop suddenly, its kinetic energy will be used to increase the temperature of the gas
We say
1/2mv^2 = ∆u
1/2mv^2 = nCv∆T
Since n = m/M
1/2mv^2 = mCv∆T/M
Making ∆T subject of the formula we have
∆T = Mv^2/2Cv
Multiple the RHS by Cp/Cp
∆T = Mv^2/2Cv *Cp/Cp
Since Y = Cp/CV
∆T = Mv^2Y/2Cp k
Since CV = R/Y - 1
We could also have
∆T = Mv^2(Y - 1)/2R k
Answer: length of B =4.00
Explanation:
for the vectors A and B and the angle between them as x.
Magnitude of the sum of A and B is given as = √(A²+B²+2ABcosx
where
Magnitude of A = 3.00
Magnitude of the sum of A and B is 5.00
5.00=√(A²+B²+2ABcos90°
5.00= √3² +b² +0
5²= 3² +b²
25=9+b²
b²= 25-9
b² = 16
b= √16
b= 4
Answer:
Explanation:
When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,
Here, n = 10
Spring constant of each spring = k
Thus,
