4. The Coyote has an initial position vector of .
4a. The Coyote has an initial velocity vector of . His position at time is given by the vector
where is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular where . Splitting up the position vector into components, we have with
The Coyote hits the ground when :
4b. Here we evaluate at the time found in (4a).
5. The shell has initial position vector , and we're told that after some time the bullet (now separated from the shell) has a position of .
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
Answer:0.253Joules
Explanation:
First, we will calculate the force required to stretch the string. According to Hooke's law, the force applied to an elastic material or string is directly proportional to its extension.
F = ke where;
F is the force
k is spring constant = 34N/m
e is the extension = 0.12m
F = 34× 0.12 = 4.08N
To get work done,
Work is said to be done if the force applied to an object cause the body to move a distance from its initial position.
Work done = Force × Distance
Since F = 4.08m, distance = 0.062m
Work done = 4.08 × 0.062
Work done = 0.253Joules
Therefore, work done to stretch the string to an additional 0.062 m distance is 0.253Joules
Answer: Light passes through the front of the eye (cornea) to the lens. The cornea and the lens help to focus the light rays onto the back of the eye (retina). The cells in the retina absorb and convert the light to electrochemical impulses which are transferred along the optic nerve and then to the brain.
Explanation:
When light rays reflect off an object and enter the eyes through the cornea (the transparent outer covering of the eye), you can then see that object. The cornea bends, or refracts, the rays that pass through the round hole of the pupil.
10/70×360°
=51.4°
hope thus helps