Answer:
Many speculation exists for why the term 'scrubbed' is used in relation to launches. <em>One of these is that during world war II, when the Air force planned bombing raids</em>. These Air force missions had their briefings done with illustration of the details of the launch and and flight drawn on chalkboards. <em>When some of mission flight takeoffs were canceled (due to bad weather or a malfunction or other reasons) the details were then scrubbed off of the chalkboard</em>. This idea of the mission <em>being scrubbed meaning that it is no longer to be followed through with</em>, proceeded to this modern day rocket and aviation launches.
Answer: The coefficient of kinetic friction is μ = 0.6
Explanation:
For an object of mass M, the weight is:
W = M*g
where g is the gravitational acceleration: g = 9.8m/s^2
And the friction force between this object and the surface can be written as:
F = W*μ
where μ is the coefficient of friction (kinetic if the object is moving, and static if the object is not moving, usually the static coefficient is larger)
In this case, the weight is:
W = 20N
And the friction force is:
F = 12N
Replacing these values in the equation for the friction force we get:
12N = 20N*μ
(12N/20N) = μ = 0.6
The coefficient of kinetic friction is μ = 0.6
Answer:
.
Explanation:
The frequency 
of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of 
. In other words, the wave would have traveled 
in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that ? The wavelength of this wave
gives the length of one wave cycle. Therefore:
.
That is: there are 
wave cycles in of this wave.
On the other hand, Because that of this wave goes through that point in each second, that wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where
represents the speed of this wave, and 
represents the wavelength of this wave.
A small vehicle with less mass and with less Kinetic Energy will require less distance to stop than a large vehicle.
Answer:
Final velocity of both goalie & puck = 0.018116 m/s
Explanation:
M1U1 + M2U2 = (M1+M2)V
70 x 0 + 0.17 x 33.5 = (70+0.17)V
V = 0.08116m/s
