Necesito hacer un póster para ciencias de estar seguro para laboratorio
1 answer:
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Answer:
W= 4.89 KJ
Explanation:
Lets take
temperature of hot water T₁ = 100⁰C
T₁ = 373 K
Temperature of cold ice T₂= 0⁰C
T₂ = 273 K
The latent heat of ice LH= 334 KJ
The heat rejected by the engine Q= m .LH
Q₂= 0.04 x 334
Q₂= 13.36 KJ
Heat gain by engine = Q₁
For Carnot engine
Q₁ = 18.25 KJ
The work W= Q₁ - Q₂
W= 18.25 - 13.36 KJ
W= 4.89 KJ
Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
