Acceleration, a = (v - u)/t
where v is the final velocity, u is the initial velocity, and t is the time.
This formula on a velocity time graph represents the slope of the graph.
Answer:
W= -2.5 (p₁*0.0012) joules
Explanation:
Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then
In adiabatic compression, work done by mixture during compression is
W= 
where f= final volume and i =initial volume, p=pressure
p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ
W= 
W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)
W=1/1-γ (p₁Vf-p₀Vi)
W= 1/1-1.40 (p₁*6/5 -p₀*0)
W= -2.5 (p₁*6/5*0.001) changing liters to m³
W= -2.5 (p₁*0.0012) joules
Answer: The reference frame of a passenger in a seat near the center of the train
Explanation:
the speed of light is the same for the passenger and the bicyclist
then the avents are simultaneous fo the passenger not for the bicyclist
the delay between the two events for the bicyclist is
Δt=Δd/vs
where
Δd= lenght of train
vs=speed of sound
the reference frame of a passenger in a seat near the center of the train
Solution:
The space and time transformations are:
x' = γ(x - vt)
t' = γ(t - vx/c²).
In the primed frame the two events are simultaneous, so that Δt' = 0. Also here Δx' = 30. In the unprimed frame Δx' = 30 = γ(Δx - v Δt).......(*)
We also have Δt' = 0 = γ(Δt - vΔx/c²)→Δx = c²Δt/v......(**)
Substituting (**) in (*): 30 = γ(c²Δt/v - vΔt))→Δt = 30/(c²/v - v) =
30/(2c - 0.5c) = 6.7 x 10^(-8)s
<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>
Answer:
U for the reaction is 15,048.58 kJ/mol of pentane.
Explanation:
Quantity of heat required = heat capacity of bomb calorimeter × temperature rise = 5.23 kJ/C × (29.8 C - 23.7 C) = 5.23 kJ/C × 6.1 C = 31.903 kJ
Moles of pentane = mass/MW
Mass = 0.1523 g
MW of pentane (C5H12) = 72 g/mol
Moles of pentane = 0.1523/72 = 0.00212 mol
U for the reaction = quantity of heat required ÷ moles of pentane = 31.903 kJ ÷ 0.00212 mol = 15,048.58 kJ/mol
