1: <span>Tides are caused by the location of the Earth, sun, and moon.
2: </span><span>Tides occur about an hour later each day.
This is because of the earth's orbit and the moons position as they move.
</span><span>
3: </span>Half as much as the moon
The moon affects the tide much more than the sun, as the moon is drastically closer to the earth than the sun.
(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.
Recall that

We have
, so that

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.
(c) The ball's average velocity is 0. Average velocity is given by
, and we know that
.
(d) The position of the ball
at time
is given by

Take the starting position to be the origin,
. Then after 6 seconds,

so the ball is 42 m away from where it started.
We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

Since the velocity is positive, the ball is still moving up the incline.
Answer:
Squats involve flexion (forward motion) and extension (backward on the way up), so would fit into the sagittal plane. Frontal plane motion would include leaning from left to right as in sidebends and lateral raises, or perhaps you might picture jumping jacks for a good image of movement along the frontal plane.
Answer:
50m [N]
Explanation:
Think of these directions as if they were on a 2D plane. (if you're talking about displacement)
If you go 32m [N] then 6m [S], it's like you're moving backwards (-). (thus, 32n-6s=26m [N])
Next, you go north again, so moving forwards (+). (thus, 26n+24n=50m [N])
Answer:

Explanation:
The radiated power can be given in terms of the wavelength as follows:

where,
Radiated Power = 1.2 x 10⁻¹⁷ W
n = no. of photons = ?
h = plank's constant = 6.625 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 500 nm = 5 x 10⁻⁷ m
t = time
Therefore,
