In general, how do you find the average velocity of any object falling in a vacuum? (Assume you know the final velocity.) Multiply the final velocity by final time. 3. Calculate : Distance, average velocity, and time are related by the equation, d = v • t A
I think it is option (C).
If the answer is helpful then mark me as brainly.
Answer: The focal length of the cornea-lens system in his eye must be LESS THAN the distance between the front and back of his eye.
Explanation:
The human eye the front part of the eye is the CORNEA. This is the tough white transparent part of the eye that helps in the refraction of light rays. While the backside of the eye is the RETINA. This is the part of the eye when images are focused.
When a normal eye is at rest, parallel rays from a distant object are focused on the retina. The ability of the eye - lens to focus points at different distances on the retina is known as accomodation. The adjustment of the eye lens to focus objects of varying distances is brought about by the ciliary muscles. The have the ability to change the shape of the eye which leads to change in focal length.
When a person with normal vision looks at a distant object at infinity, the lens brings parallel rays to focus on the retina. Thus, the furthest point which the eye can see distinctly is called the far point of the eye and it's infinity for a normal eye. But Joe was able to focus his eye on the tree, meaning that the tree was within his near point. This is the nearest point at which an object is clearly seen. Therefore, when the effective focal length of the cornea-lens system changes, it changes the location of the image of any object in one's field of view.
Answer:
Explanation:
from the question we have the following:
distance between Sacramento and los angles = 400 miles
speed of car A = 60 mph
start time of car A = 11 am
speed of car B = 75 mph
start time of car B = 12 pm
distance of Fresno from Los Angeles = 150 miles
- To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
- Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
- From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
- Distance covered by car A = speed x time(t) = 60 x t = 60t
- Distance covered by car B = speed x time(t) = 75 x t = 75t
- 60t + 75t = 340 miles
- 135t = 340
- t = 2.51 hours
- Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
- Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
- 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.
