For this we will use kinematic equations.
We need the time of flight to compute the range.
With the given information, the most useful formula is:
Δy=v0yt−g2t2
Because our equation is taking only vertical motion into account,
v0y=30sin(45)
Because we are looking for the range of the object,
Δy=0
This leaves us with only one unknown: t
0=30sin(45)t−g2t2
g2t2=30sin(45)t
g2t=30sin(45)
t=60gsin(45)
For horizontal distance, the formula we need is:
Δx=v0xt
Because we are taking into account the horizontal range,
v0x=30cos(45)
In substituting our derived values into the equation we can see that:
Δx=30cos(45)∗60gsin(45)
Range=30cos(45)∗60gsin(45)
C. Because you’re stating a list of 3
Answer:
6.47x10-5
Explanation:
I just did it on my I ready test
The matchup statement associated with rods or cones are:
<h3>Rods:</h3>
- involved with night vision
- smooth plasma membrane surrounding the intramembranous discs
- uses rhodopsin
- used in scotopic vision
- absorption peak occurs at 500 nm
<h3>Cones:</h3>
- highly concentrated in and around the macula
- involved with color vision
- infolded plasma membrane surrounds each disc
- uses photopsin
- absorption peaks occur at 420, 531, and 558 nm
<h3>What are the roles of rods and cones in the eye?</h3>
The Rods are known to help in regards to vision at low light levels or scotopic vision.
The cones are known to help or function in regards to vision at the point of higher light levels or what we call photopic vision.
Hence the matchup above are correct.
Learn more about rods and cones from
brainly.com/question/930059
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