Answer:
The thrust is 
Explanation:
Given that,
Mass of gas, 
The rate at which the gas is expelling, 
We need to find the thrust produced by the gas.
We know that force is equal to the rate of change of momentum. So,
Also, p = mv
So,
So, the thrust is
Answer:
182 to 3 s.f
Explanation:
Workdone for an adiabatic process is given as
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
where γ = ratio of specific heats. For carbon dioxide, γ = 1.28
For an adiabatic process
P₁V₁ʸ = P₂V₂ʸ = K
K = P₁V₁ʸ
We need to calculate the P₁ using ideal gas equation
P₁V₁ = mRT₁
P₁ = (mRT₁/V₁)
m = 2.80 g = 0.0028 kg
R = 188.92 J/kg.K
T₁ = 27°C = 300 K
V₁ = 500 cm³ = 0.0005 m³
P₁ = (0.0028)(188.92)(300)/0.0005
P₁ = 317385.6 Pa
K = P₁V₁¹•²⁸ = (317385.6)(0.0005¹•²⁸) = 18.89
W = K(V₂¹⁻ʸ - V₁¹⁻ʸ)/(1 - γ)
V₁ = 0.0005 m³
V₂ = 2.10 dm³ = 0.002 m³
1 - γ = 1 - 1.28 = - 0.28
W =
18.89 [(0.002)⁻⁰•²⁸ - (0.0005)⁻⁰•²⁸]/(-0.28)
W = -67.47 (5.698 - 8.4)
W = 182.3 = 182 to 3 s.f
The object is moving, so at different times, it has different displacement. I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.
Displacement is the distance and the direction FROM (the position at the beginning) TO (the position at the end).
At the beginning ... time=0 ... the position is 1 meter.
At the end ... time=5 ... the position is zero.
The distance FROM the beginning TO the end is (zero - 1m) . That's <em>-1m </em>.
Here in order to find out the distance between two planes after 3 hours can be calculated by the concept of relative velocity
here
speed of first plane is 700 mi/h at 31.3 degree
speed of second plane is 570 mi/h at 134 degree
now the relative velocity is given as
now the distance between them is given as
so the magnitude of the distance is given as
miles
so the distance between them is 2985.6 miles
