Today's radio consists of an antenna, printed circuit board, resistors, capacitors, coils and transformers, transistors, integrated circuits, and a speaker. All of these parts are housed in a plastic case. An internal antenna consists of small-diameter insulated copper wire wound around a ferrite core.
Liquid Hydrogen is the fuel used by rockets.
Explanation:
- Liquid hydrogen which can be chemically denoted as "
" is often considered as the significant fuels for rocket.
- However rocket in its lower stages uses fuels such as Kerosene and oxygen where as in the higher stages such as second and third stages it uses liquid hydrogen.
- Liquid hydrogen is known to easily cool the nozzle and then also other parts of the rocket before mixing with the oxidizer such as the oxygen.
- Thus liquid hydrogen helps in preventing nozzle erosion and also reduces combustion chamber.
- Liquid hydrogen one the other hand is very expensive as 384,071 gallons of it will cost approximately $376,389.58
.
Thus liquid hydrogen is effectively used as a fuel for rocket.
Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
True the use many abstract power