Answer:
The acceleration of the sprinter is 1.4 m/s²
Explanation:
Hi there!
The equation of position of the sprinter is the following:
x = x0 + v0 · t + 1/2 · a · t²
Where:
x = position of the sprinter at a time t.
x0 = initial position.
v0 = initial velocity.
t = time.
a = acceleration.
Since the origin of the frame of reference is located at the starting point and the sprinter starts from rest, then, x0 and v0 are equal to zero:
x = 1/2 · a · t²
At t = 9.9 s, x = 71 m
71 m = 1/2 · a · (9.9 s)²
2 · 71 m / (9.9 s)² = a
a = 1.4 m/s²
The acceleration of the sprinter is 1.4 m/s²
According to the question, the object is placed at 2F
The ray diagram is shown in the figure attached.
According to the figure:
Object AB is at 2F₁
First, we draw a ray parallel to principal axis.
So, it passes through focus after refraction.
We draw another ray which passes through optical center.
So, the ray will go through without any deviation.
Where both refracted rays meet is point A' and the image formed is A'B'
This image is formed at 2F₂
We can say that:
- Image is real.
- Image is inverted.
- Image is exactly the same size as the object.
The answer is C, a peer group.
The rock it traveling really, really fast.
It is hard to exactly determine how fast bc u need the height of the cliff and how big the rock is.
Hope this helps and can I get brainliest answer!
