Answer:
c
Explanation:
Hello!
To solve this problem we need to know the velocity and position of the train as a fucntion of time and acceleration. Since the train started from rest, these are the formulas:
v = at
x = (1/2)a*t^2
From the first equation we can know the time as a division between the velocity and the acceleration:
t = (v/a)
Replacing this value in the second equation we get:
x = (1/2)*a*(v/a)^2 = v^2/(2*a)
Solving for a:
a = (1/2)*(v^2/x)
Now, we know that when x=5.6km =5600 m, the velocity of the train is v =42 m/s
Therefore:
a = (1/2)*(42^2/5600) m/s^2 = 0.1575 m/s^2
So, the answer is c, the acceleration of the train during the first 5.6 km is colse to 0.16m/S^2
M1V1 + M2V2 = M1V1' + M2V2'
where:
M1 is the mass of the large marble = 0.05 kg
V1 is the initial velocity of the large marble = 0.6 m/sec
M2 is the mass of the small marble = 0.03 kg
V2 is the initial velocity of the small marble = 0 m/sec (marble is at rest)
V1' is the final velocity of the large marble = -0.2 m/sec
V2' is the final velocity of the small marble that we want to calculate
Substitute with the givens in the above equation to get V2' as follows:
M1V1 + M2V2 = M1V1' + M2V2'
(0.05)(0.6) + (0.03)(0) = (0.05)(-0.2) + 0.03V2'
0.03 = -0.01 + 0.03V2'
0.03V2' = 0.03+0.01 = 0.04
V2' = 0.04/0.03
V2' = 1.334 m/sec
So we want to know what is the distance d of the object from the lens if the height of the object is h=6 cm, focal length of the lens is f=5 cm and the distance d=15 cm is the distance of the object from the lens. From the formula for the convex lens 1/f=(1/D + 1/d) where D is the distance of the image from the lens we can get D after solving for D: 1/D=(1/f) - (1/d),
1/D=(1/5)-(1/15)=0.2-0,06667=0.13333 so f=1/0.13333=7.500187 cm. Rounded to the nearest hundredth D=7.50 cm. That is very close to 7.69 cm so the correct answer is the third one.