Un bombero alejado d = 31.0 m de un edificio en llamas dirige un chorro de agua desde una manguera contra incendios a nivel del
1 answer:
Answer:
El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.
Explanation:
El chorro de agua exhibe un movimiento parabólico, dado que este tiene una inclinación inicial y la única aceleración es debida a la gravitación terrestre. Las ecuaciones cinemáticas que modelan el fenómeno son:
Distancia horizontal (en metros)
Donde:
- Posición horizontal inicial, medida en metros.
- Tiempo, medido en segundos.
- Velocidad inicial, medida en metros por segundo.
- Angulo de inclinación del chorro de agua, medido en grados sexagesimales.
Distancia vertical (en metros)
Donde:
- Posición vertical inicial, medida en metros.
- Constante gravitacional, medida en metros por segundo al cuadrado.
Partiendo de la primera ecuación, se despeja el tiempo:
Si ,
,
and
, entonces:
La altura máxima se calcula por sustitución directa de términos en la segunda ecuación. Si ,
,
y
, entonces:
El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.
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In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.
2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.
<span>A.Wavelengths measured would match the actual wavelengths emitted.
In fact, the stars are not moving relative to Earth, so there is no shift in the measured wavelength.
2) In this second case, the correct answer is
</span><span>A.Wavelengths measured would be shorter than the actual wavelengths emitted.
</span>in fact, since the stars in this case are moving towards the Earth, then apparent frequency of their emitted light will be larger than the actual frequency, because of the Doppler effect, according to the formula:
where f0 is the actual frequency, f' the apparent frequency, c the speed of light and vs the velocity of the source (the stars) relative to the obsever (Earth). Vs is negative when the source is moving towards the observer, so the apparent frequency f' is larger than the actual frequency f0. But the wavelength is inversely proportional to the frequency, so the apparent wavelength will be shorter than the actual wavelength.
Answer:
a
The orbital speed is
b
The escape velocity of the rocket is
Explanation:
Generally angular velocity is mathematically represented as
Where T is the period which is given as 1.6 days =
Substituting the value
At the point when the rocket is on a circular orbit
The gravitational force = centripetal force and this can be mathematically represented as
Where G is the universal gravitational constant with a value
M is the mass of the earth with a constant value of
r is the distance between earth and circular orbit where the rocke is found
Making r the subject
The orbital speed is represented mathematically as
Substituting value
The escape velocity is mathematically represented as
Substituting values
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s