Answer:
Explanation:
E = σ/ε = (F/A) / (ΔL/L)
E = (mg/(πd²/4) / (ΔL/L)
E = (4mg/(πd²) / (ΔL/L)
E = 4Lmg/(πd²ΔL)
E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)
E = 1.35 x 10⁹ Pa or 1.35 GPa
Answer:
When a candle was blown out, the flame stops immediately but the wick and the wax are both still hot, so pyrolysis continues for a few seconds. Explanation:
Answer:
Q=185.84C
Explanation:
We have to take into account the integral
![Q=\int \rho dV](https://tex.z-dn.net/?f=Q%3D%5Cint%20%5Crho%20dV)
In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4
![Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx](https://tex.z-dn.net/?f=Q%3D%5Cint_%7B-2%7D%5E2%5Cint_%7B-%5Csqrt%7B4-x%5E2%7D%7D%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx)
but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!
A=atomic mass
Z=atomic number (nº of protons).
N=neutons.
A=Z+N
Data:
A=39
Z=19
A=Z+N
39=19+N
N=39-19=20.
It therefore has 20 neutrons in its nucleus.
Answer:
Explanation:
a) R = V/I = 0.7/0.022 = 32Ω
b) R = V/I = 0.6/0.005 = 120Ω
c) No...Current change is not linear with voltage change.