Incomplete question as there is so much information is missing.The complete question is here
A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?
Answer:
Distance traveled=240 m
Explanation:
Given data
Initial velocity of car v₀=0 m/s
Final velocity of car vf=24 m/s
Distance traveled by car S=120 m
To find
Distance does the traffic travel
Solution
To find the distance first we need to find time, for time first we need acceleration
So

As we find acceleration.Now we need to find time
So

Now for distance
So

Answer:
The separation distance between the slits is 16710.32 nm.
Explanation:
Given that,
Wavelength = 641 nm
Angle =4.4°
(a). We need to calculate the separation distance between the slits
Using formula of young's double slit


Where, d = the separation distance between the slits
m = number of order
=wavelength
Put the value into the formula



Hence, The separation distance between the slits is 16710.32 nm.
Answer:
100m
Explanation:
Find the acceleration first
Then plot a simple speed time graph
Answer
225 meters.
Explanation:
x=x0+30t-(1/2)(1.5)t^2
x=0+30(10)-(1/2)(1.5)(10)^2
x=300-75
x=225
Ball C has the greatest momentum and ball B has the least amount of momentum.