As per Einstein's theory of special relativity, the light speed in a vacuum remains constant regardless of the observer's speed. Therefore, the response should be A) 0.1c (one-tenth the speed of light)
The principle of momentum conservation<span> is a key law in the field of physics. It asserts that the </span>momentum<span> within a system remains unchanged unless there are </span>external forces influencing the system. In the case of two balls, each weighing 0.5 kg, colliding on a pool table<span>, this principle does not hold because external forces acted upon the balls during the collision. </span>
Answer:
1.5 m/s²
Explanation:
Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.
Considering the forces perpendicular to the incline:
∑F = ma
N − mg cos θ = 0
This gives us N = mg cos θ
Next, examining the forces parallel to the incline:
∑F = ma
mg sin θ − Nμ = ma
Substituting for N yields:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g cos θ μ = a
hence a = g (sin θ − μ cos θ)
If we set θ = 23° and μ = 0.26:
a = 9.8 (sin 23 − 0.26 cos 23)
this results in a = 1.48
When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².
Definamos h como la distancia que hay desde el borde del pozo hasta la superficie del agua (en metros).
Consideremos la gravedad g como 9.8 m/s² y despreciemos la resistencia del aire.
La velocidad inicial vertical del guijarro es nula.
Ya que el guijarro impacta el agua tras 1.5 segundos, entonces:
h = 0.5 * (9.8 m/s²) * (1.5 s)² = 11.025 m
Resultado: 11.025 m
