Answer: It should be A or the very left red circle that you can click on
Explanation: Because when the wind is moving downward and the earth is spinning the spot the wind ends up will never be directly down from where it was to begin with
D.Radiant energy does not require a medium through which to travel.
Answer is 6 tires.
This is a projectile question.
First make sure units are consistent - express speed in m/s.
20 km/h = 20000m / 3600 s = 5.56 m/s
Assume the takeoff point of the ramp is at ground level (height, h, = 0m). We need to determine how long Joe is in the air, and use that time to calculate the horizontal distance he traveled.
Joe is traveling 5.56 m/s on a ramp angled at 20 degrees. There are vertical and horizontal components to his speed:
Vertical speed = 5.56sin20 = 1.90 m/s
Horizontal speed = 5.56cos20 = 5.22 m/s
An easy way to proceed is to calculate the time it takes for Joe’s vertical speed to reach 0m/s - this represents the time when Joe is at his maximum height and is therefore halfway through the trip. Double whatever time this is to find the total time of the trip. Remember he is decelerating due to gravity:
Time to peak:
a = Δv / Δt
-9.8 = -1.9 / Δt
Δt = 0.19s
Total trip time:
0.19 x 2 = 0.38s
Now that we have the total tome Joe is in the air, we can find the horizontal distance he traveled:
v = d / t
5.22 = d / 0.38
d = 1.98m
Now divide this total distance by the length of an individual tire to find the number of tires he will clear:
1.98 / 0.3 = 6.6 tires
Therefore he can jump 6 tires safely (he will land in the middle of the 7th tire).
Lots of steps I know but just try to think of the situation and keep track of the vertical and horizontal things!
Answer:
The excess charge has distributed itself evenly over the outside surface of the sphere.
Explanation:
Since the hollow sphere is a conductor, it has free electrons that can move about within the sphere. In this light, an excess charge, the like charges repels each other, therefore ensuring that charges are spread as far apart as possible. There is therefore an evenly distributed charge on the outside surface of the sphere.
Explanation:
We need to apply the conservation law of linear momentum to two dimensions:
Let 
= momentum of the 1st ball
= momentum of the 2nd ball
In the x-axis, the conservation law can be written as
or
Since we are dealing with identical balls, all the m terms cancel out so we are left with
Putting in the numbers, we get
In the y-axis, there is no initial y-component of the momentum before the collision so we can write
or
Taking the ratio of the sine equation to the cosine equation, we get
or
Solving now for 
,
