Answer:
La aceleración necesaria para detener el avión es - 10.42 m/s².
Explanation:
Un movimiento uniformemente acelerado (M.U.A) es aquél cuya aceleración es constante y la velocidad de un objeto cambia a medida que el movimiento evoluciona.
Siendo la aceleración "a" el cambio de velocidad al tiempo transcurrido en un punto A a B, la velocidad inicial la velocidad que tiene un cuerpo al iniciar su movimiento en un período de tiempo y la velocidad final la velocidad que tiene un cuerpo al finalizar su movimiento en un período de tiempo, entonces en M.U.A se cumple:
Vf² - Vo² = 2*a*d
donde:
- Vf: Velocidad final
- Vo: Velocidad inicial
- a: Aceleración
- d: Distancia recorrida
En este caso:
- Vf: 0 m/s, porque el avión se detiene
- Vo: 50 m/s
- a: ?
- d: 120 m
Reemplazando:
(0 m/s)² - (50 m/s)² = 2*a*120 m
Resolviendo:
a= - 10.42 m/s²
<u><em>La aceleración necesaria para detener el avión es - 10.42 m/s².</em></u>
Answer:
b) The star is moving away from us.
Explanation:
If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.
#1. A. Waxing crescent.
#2. 1.
#3. C.
#4. C.
Answer:
a) Em₀ = 42.96 104 J
, b) 
= -2.49 105 J
, c) vf = 3.75 m / s
Explanation:
The mechanical energy of a body is the sum of its kinetic energy plus the potential energies it has
Em = K + U
a) Let's look for the initial mechanical energy
Em₀ = K + U
Em₀ = ½ m v2 + mg and
Em₀ = ½ 50.0 (1.20 102) 2 + 50 9.8 142
Em₀ = 36 104 + 6.96 104
Em₀ = 42.96 104 J
b) The work of the friction force is equal to the change in the mechanical energy of the body
= Em₂ -Em₀
Em₂ = K + U
Em₂ = ½ m v₂² + m g y₂
Em₂ = ½ 50 85 2 + 50 9.8 427
Em₂ = 180.625 + 2.09 105
Em₂ = 1,806 105 J
= Em₂ -Em₀
= 1,806 105 - 4,296 105
= -2.49 105 J
The negative sign indicates that the work that force and displacement have opposite directions
c) In this case the work of the friction going up is already calculated in part b and the work of the friction going down would be 1.5 that job
We have that the work of friction is equal to the change of mechanical energy
= ΔEm
= Emf - Emo
-1.5 2.49 10⁵ = ½ m vf² - 42.96 10⁴
½ m vf² = -1.5 2.49 10⁵ + 4.296 10⁵
½ 50.0 vf² = 0.561
vf = √ 0.561 25
vf = 3.75 m / s
