Answer:
Explanation:
Given:
U1 = 1.6 m/s
U2 = -1.1 m/s
M1 = 1850 kg
M2 = 1400 kg
V1 = 0.27 m/s
Using momentum- collision equation,
M1U1 + M2U2 = M1V1 + M2V2
1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2
1420 = 499.5 + 1400V2
V2 = 0.6575 m/s
B.
KE = 1/2 × MV^2
KEa1 + KEa2 = KEb1 + KEb2
Delta KE = KE2 - KE1
KEa1 = 2368 J
KEb1 = 847 J
KEa2 = 67.433 J
KEb2 = 302.6 J
KE1 = KEa1 + KEb1
= 3215 J
KE2 = 370.033 J
Delta KE = -2845 J.
Answer:
6.6 atm
Explanation:
Using the general gas law
P₁V₁/T₁ = P₂V₂/T₂
Let P₂ be the new pressure
So, P₂ = P₁V₁T₂/V₂T₁
Since V₂ = 2V₁ , P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius
T₂ = 273 + 2t (since its Celsius temperature doubles).
Substituting these values into the equation for P₂, we have
P₂ = P₁V₁(273 + 2t)/2V₁(273 + t)
P₂ = 12(273 + 2t)/[2(273 + t)]
P₂ = 6(273 + 2t)/(273 + t)]
assume t = 30 °C on a comfortable spring day
P₂ = 6(273 + 2(30))/(273 + 30)]
P₂ = 6(273 + 60))/(273 + 30)]
P₂ = 6(333))/(303)]
P₂ = 6.6 atm
Usually describes a system by a set of variables in a set of equations established relationships between the variables and variables maybe of many types real or integer numbers Boolean values of strings for example
For a storm to be a blizzard, the wind must be at least 35 miles per hour. This is just one criteria for considering a storm to be a blizzard. The wind speed of 35 miles per hour should reduce the visibility to less than 400 meters. The last criteria is that the storm must continue for a time frame of at least 3 hours.
