Answer:
The maximum height the box will reach is 1.72 m
Explanation:
F = k·x
Where
F = Force of the spring
k = The spring constant = 300 N/m
x = Spring compression or stretch = 0.15 m
Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N
Mass of box = 0.2 kg
Work, W, done by the spring = 
and the kinetic energy gained by the box is given by KE = 
Since work done by the spring = kinetic energy gained by the box we have
= therefore we have v = 
= 
= 
= 5.81 m/s
Therefore the maximum height is given by
v² = 2·g·h or h = 
= 
= 1.72 m
Answer:
5235.84 kg
Explanation:
There is one theorem - whose proof I will never remember without having to drag calculus in there - that says that the variation of momentum is equal to the force applied times the time the application last.
As long as the engine isn't ejecting mass - at this point it's a whole new can of worm - we know the force, we know the variation in speed, time to find the mass. But first, let's convert the variation of speed in meters per second. The ship gains 250 kmh, 
;
Answer:
B
Explanation:
if no force acts upon it it will just continue moving
hope this helps!! have a wonderfull day!
Answer:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Explanation:
El gasto es el flujo volumétrico de gasolina (
), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:
(1)
Donde:
- Diámetro de la manguera, medido en pies.
- Velocidad medida de salida, medida en pies por segundo.
Si sabemos que 
y 
, entonces el gasto de gasolina es:
El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.
Answer:They will go off in opposite directions with the same force.
Explanation:
Newton's Third Law states that for every action, there is an equal but opposite reaction. If two people are standing on a scooter and push off each other (following the Law), it should come to the conclusion that they could go off in opposite with the same force.
