Si un electrón recorre el acelerador lineal de Stanford de 2 millas de longitud a 99% de la velocidad de la luz, ¿Cuál es la lon
1 answer:
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Answer:
a) t=1s
y = 10.1m
v=5.2m/s
b) t=1.5s
y =11.475 m
v=0.3m/s
c) t=2s
y =10.4 m
v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Explanation:
Conceptual analysis
We apply the free fall formula for position (y) and speed (v) at any time (t).
As gravity opposes movement the sign in the equations is negative.:
y = vi*t - ½ g*t2 Equation 1
v=vit-g*t Equation 2
y: The vertical distance the ball moves at time t
vi: Initial speed
g= acceleration due to gravity
v= Speed the ball moves at time t
Known information
We know the following data:
Vi=15 m / s
t=1s ,1.5s,2s
Development of problem
We replace t in the equations (1) and (2)
a) t=1s
=15-4.9=10.1m
v=15-9.8*1 =15-9.8 =5.2m/s
b) t=1.5s
=22.5-11.025=11.475 m
v=15-9.8*1.5 =15-14.7=0.3m/s
c) t=2s
= 30-19.6=10.4 m
v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)
Answer:
Explanation:
Attached is the solution
