Answer:
The angle for the forward Mach line is 19.47°
The angle for the rearward Mach line is 5.21°
Explanation:
From table A-1 (Modern Compressible Flow: with historical perspective):
(M₁ = 3)
If Po₁ = Po₂
Table A-1:
Table A-5:
v₁ = 49.76°
μ₁ = 19.47°
v₂ = 60.55°
μ₂ = 16°
θ = 60.55 - 49.76 = 10.79°
The angle for the forward Mach line is:
μ₁ = 19.47°
The angle for the rearward Mach line is:
θr = μ₂ - θ = 16 - 10.79 = 5.21°
Answer:
energy is equal to 1000 J
Explanation:
When the jumper is in the tent, he has a given height, this height gives him a gravitational potential energy, which forms his initial mechanical energy of 1000 J. After jumping, this energy is converted into elastic energy of the rope plus a remainder of potential energy gravitational, it does not reach the ground, but as the friction is negligible the total mechanical energy is conserved, therefore its energy is equal to 1000 J
This is a case of energy transformation, but the total value of mechanical energy does not change
The equation of D = m/V
Where D = density
m = mass
and V = volume
We are solving for V, so with the manipulation of variables we multiply V on both sides giving us
V(D) = m
now we divide D on both sides giving us
V = m/D
We know our mass which is 600g and our density is 3.00 g/cm^3
so
V = 600g/3.00g/cm^3 = 200cm^3 or 200mL
a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL
If you wish to change it to L, you'd have to convert.
Answer:
La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.
Explanation:
El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.
La distancia recorrida, x
, por un móvil que tiene un MRU con un velocidad v durante el intervalo de tiempo t es:
x= x0 + v*t
donde x0 es la posición inicial.
En este caso:
Reemplazando:
x= 22 m + 5 m/s* 30 s
Resolviendo:
x= 22 m + 150 m
x= 172 m
<u><em>La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.</em></u>
