An ideal gas trapped inside a thermally isolated cylinder expands slowly by pushing back against a piston. The temperature of th
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Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land
Answer:
n = 4 x 10¹⁸ photons
Explanation:
First, we will calculate the energy of one photon in the radiation:
where,
E = Energy of one photon = ?
h = Plank's Constant = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength of radiation = 567 nm = 5.67 x 10⁻⁷ m
Therefore,
E = 3.505 x 10⁻¹⁹ J
Now, the number of photons to make up the total energy can be calculated as follows:
<u>n = 4 x 10¹⁸ photons</u>