Answer:
the ship's energy is greater than this and the crew member does not meet the requirement
Explanation:
In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship
W =∫ F dx = ΔK
Let's replace
∫ (α x³ + β) dx = ΔK
α x⁴ / 4 + β x = ΔK
Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J
x (α x³ + β) = 
- K₀
= K₀ + x (α x³ + β)
Assuming that the low limit is x = 0, measured from the cargo hangar
Let's calculate
= 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)
Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)
Kf = 2.7 10¹¹ - 1.1475 10¹¹
Kf = 1.55 10¹¹ J
In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement
We evaluate the kinetic energy if the System is well calibrated
W = x F₀ = –K₀
= K₀ + x F₀
We calculate
= 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶
= (2.7 -2.625) 10¹¹
= 7.5 10⁹ J
El factor mas importante para el exceso de peso es un exceso de energía creada por una alimentación excesiva
El peso de un cuerpo es definido por la relación entre la energía requerida para los procesos vitales del cuerpo, sus actividades físicas diarias y la energía suministra en forma de alimentos.
Cuando estos dos parámetros están en balance el peso es estable, pero cuando la cantidad de alimentos aumenta o el valor energético de los mismo aumenta se tiene un exceso de energía que el cuerpo almacena en forma de grasa corporal, este el el factor mas importante para el exceso de peso.
En conclusión el factor mas importante para el exceso de peso es un exceso de energía por una alimentación excesiva
aprende mas acerca del peso corporal aquí:
brainly.com/question/13032223
Answer:
The answer to your question is letter B.
Explanation:
To answer this question, we must remember the third law of motion of Newton that states that For every action, there is an equal and opposite reaction.
Then, if the action force is 40 N to the right, the reaction force must be 40 N to the left.
Refer to the diagram shown below.
We want to find y in terms of d, φ and θ.
By definition,
Therefore
y = x tan(θ) (1)
y = (x - d) tan(φ) (2)
Equate (1) and (2).
![(x - d) \, tan(\phi) = x \, tan(\theta) \\ x[tan(\phi) - tan(\theta)] = d \, tan(\phi) \\ x= \frac{d tan(\phi)}{tan(\phi)-tan(\theta)}](https://tex.z-dn.net/?f=%28x%20-%20d%29%20%5C%2C%20tan%28%5Cphi%29%20%3D%20x%20%5C%2C%20tan%28%5Ctheta%29%20%5C%5C%20x%5Btan%28%5Cphi%29%20-%20tan%28%5Ctheta%29%5D%20%3D%20d%20%5C%2C%20tan%28%5Cphi%29%20%5C%5C%20x%3D%20%5Cfrac%7Bd%20tan%28%5Cphi%29%7D%7Btan%28%5Cphi%29-tan%28%5Ctheta%29%7D%20)
From (1), obtain the required expression for y.
Answer:
