Susan works in the research and development department. She has recently purchased a large high-speed external drive and has att
1 answer:
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To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,
Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective
Replacing with our values
Therefore the focal length of th eobjective lenses is 27.75cm
Answer:
1) 3.66 s
2) 124.44 m
3) 3.12 s
Explanation:
Let's start by first listing down the information in the question.
Red Car : 34 m/s
Blue Car: 28 m/s
Distance between them : 22 m
The difference in speed between the cars is: 34 - 28 = 6 m/s
This means that the red car is catching up to the blue car at a speed of 6 m/s.
1) We can solve this by just dividing the distance by the difference in speed. This becomes:
Thus it takes 3.66 seconds for the red car to catch up to the blue car.
2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.
This becomes:
Thus the red car travels 124.44 m before catching up to the blue car.
3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.
We can use the motion equation :
Here s = 22 m
We can take u as the difference in speed. u = 6 m/s
Acceleration a = (2/3) m/s^2
Substituting the these into the equation we get:
Solving this for the variable 't' using the quadratic formula we get the following two answers:
t1 = 3.12 s
t2 = - 21.12 s
Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.