Answer:
p = 6.64 cm
Explanation:
For this exercise we use the equation of the constructor
where f is the focal length, p and q are the distance to the object and the image, respectively
They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image
1 / p = 1 / 2.2 - 1/3.29
1 / p = 0.15059
p = 6.64 cm
therefore the farthest distance from the object is 6.64 c
The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin
θ.
The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.
If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.
Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
Learn more here; brainly.com/question/12870645
Answer:
Explanation:
graph would be a straight line from (0, 0) to (400, 8)
Plot points are
PE = mgh
50(0) = 0 J
50(2) = 100 J
50(4) = 200 J
50(6) = 300 J
50(8) = 400 J
