Answer:
A) m = 1600 kg
, B) p = 3.2 10⁴ kg m / s
, C) Δp = - 3.2 10⁴ kg m / s
D) t = 51.6 s
Explanation:
A) They indicate the weight of the vehicle W = 15680 N
weight is the attraction of the earth on a body
W = m g
m = W / g
m = 15680 / 9.8
m = 1600 kg
B) The momentum is defined by
p = m v
p = 1600 20
p = 3.2 10⁴ kg m / s
C) the moment change is
Δp = p_{f} -p₀
as the vehicle stops the final speed is zero and therefore the moment is zero
ΔP = 0 - p₀
Δp = - 3.2 10⁴ kg m / s
D) Let's use Newton's second law
F = m a
a = F / m
a = 6.2 102/1600
a = 0.3875 m / s²
Now we can use kinematics, note that as the vehicle stops, the relationship must be opposite to the speed
v = v₀ - a t
v = 0
0 = v₀ - a t
t = v₀ / a
t = 20 / 0.3875
t = 51.6 s
Answer:
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
Explanation:
You can find the field using Gauss's Law:
the surface S is an "infinite long" cylinderr of radio r.
![\int\ {E.} \, dS = E\int\ dS=E.S=E2\pi rL](https://tex.z-dn.net/?f=%5Cint%5C%20%7BE.%7D%20%5C%2C%20dS%20%3D%20E%5Cint%5C%20dS%3DE.S%3DE2%5Cpi%20rL)
![Qin=\lambda L](https://tex.z-dn.net/?f=Qin%3D%5Clambda%20L)
E(r)=![\frac{\lambda }{2\pi \epsilon} . \frac{1}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Clambda%20%7D%7B2%5Cpi%20%5Cepsilon%7D%20.%20%5Cfrac%7B1%7D%7Br%7D)
λ=-92.0 μC/m, ε=8.85.10^-12
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
The answer is B. natural gas
To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.
The energy fluency is given by the equation
![\Psi=4RcE\pi](https://tex.z-dn.net/?f=%5CPsi%3D4RcE%5Cpi)
Where
The energy fluency
c = Activity of the source
r = distance
E = electric field
In the other hand we have the equation for current in materials, which is given by
![I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}](https://tex.z-dn.net/?f=I%3D%20I_0%20e%5E%7B-%5Cmu_%7Bh20%7DX_%7Bh2o%7D%7D%20e%5E%7B-%5Cmu_%7BFe%7DX_%7BFe%7D%7D)
Then replacing our values we have that
![I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}](https://tex.z-dn.net/?f=I%20%3D%201%2A10%5E%7B-3%7D%20%2A%203.3%2A10%5E%7B10%7D%20%2A%20e%20%5E%7B-0.06%2A1.1%7D%20e%5E%7B-0.058%2A7.861%7D)
![I = 1.3*10^7 Bq](https://tex.z-dn.net/?f=I%20%3D%201.3%2A10%5E7%20Bq)
We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.
Now the energy fluency would be,
![\Psi = \frac{cE}{4\pir^2}](https://tex.z-dn.net/?f=%5CPsi%20%3D%20%5Cfrac%7BcE%7D%7B4%5Cpir%5E2%7D)
![\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}](https://tex.z-dn.net/?f=%5CPsi%20%3D%20%5Cfrac%7B1.3%2A10%5E7%2A2%2A1.25%7D%7B4%5Cpi%2A11%5E2%7D)
![\Psi = 2.14*10^4 MeV/cm^2.s](https://tex.z-dn.net/?f=%5CPsi%20%3D%202.14%2A10%5E4%20MeV%2Fcm%5E2.s)
The uncollided flux density at the outer surface of the tank nearest the source is ![\Psi = 2.14*10^4 MeV/cm^2.s](https://tex.z-dn.net/?f=%5CPsi%20%3D%202.14%2A10%5E4%20MeV%2Fcm%5E2.s)
You have to change the properties of the medium that the sound waves traveling through