Average speed = (distance covered) / (time to cover the distance)
= (30km + 40 km) / (25min + 35min)
= (70 km) / (60 min)
= 70 km/hour
Answer: "One object speeds up before it slows to a stop
"
(the top one)
Explanation:
Ok, first a little recall on how to read this type of graph.
If the points are far apart, the object is moving fast.
If the points are close together, the object is moving slow.
If the distance between the points changes then the velocity of the object changes, which means that the object is accelerated.
If we have a lot of points clustered in one location, then the object is not moving.
We can see:
The top object starts slow, then it increments the speed, then it slows down again, and then it comes to stop.
The bottom object starts fast, and it slows down.
then:
"One object speeds up before it slows to a stop
"
This describes the motion of the top object, this is the only correct option that describes one of the graphs.
initial velocity = 0 = v₍i₎
final velocity = ? = v₍f₎
t = 12 sec
Acceleration = 4m/s²
First we have to find the distance d, for this we use the
formula,
D = v₍i₎t + 1/2at²
D = 0(12) + ½ (4)(12)²
Distance = d = 288 m
Now to find the Vf use the formula,
V₍f₎² = v₍i₎² + 2ad
V₍f₎² = (0)2 + 2(4)(288)
V₍f₎² = 2304
V₍f₎ = 48 m/s
so the velocity at the end of 12 sec is 48 m/s
Answer:
La frecuencia será la misma en los dos medios, y en el vacio, no varia.
b) Vacío : n = 3,684 10 m 1,33 = 4 900 Å ,3161 10 m 55,1 n
,6 105 10 Hz
3,161 10 m
v 1,93 10 m/s 193548,38 km/s =
1,55
10 3 m/s
n
c
Vidrio v: =
,6 107 10 Hz
3,684 10 m
v 2,25 10 m/s 225 000km/s =
33,1
10 3 m/s
n
c
a
