Answer:
KE = 10530 J or 10.53 KJ
Explanation:
The formula for kinetic energy is KE = 1/2 mv^2
Let's apply the formula:
KE = 1/2 mv^2
KE = 1/2 (65kg) (18m/s)^2
KE = 10530 J or 10.53 KJ
<span>Now that you know the time to reach its maximum height, you have enough information to find out the initial velocity of the second arrow. Here's what you know about it: its final velocity is 0 m/s (at the maximum height), its time to reach that is 2.8 seconds, but wait! it was fired 1.05 seconds later, so take off 1.05 seconds so that its time is 1.75 seconds, and of course gravity is still the same at -9.8 m/s^2. Plug those numbers into the kinematic equation (Vf=Vi+a*t, remember?) for 0=Vi+-9.8*1.75 and solve for Vi to get.......
17.15 m/s</span>
50% of the moon is always illuminated, however during it's quarter phase means that we only see a quarter of what's really lit up. So it LOOKS like the moon is only 25% lit and 75% dark, it's truly 50/50. We only see that 25% since we can see it from one angle.
Answer:
The maximum speed at which the car can safety travel around the track is 18.6m/s.
Explanation:
Since the car is in circular motion, there has to be a centripetal force 
. In this case, the only force that applies for that is the static frictional force 
between the tires and the track. Then, we can write that:
And since 
and 
, we have:
Now, if we write the vertical equation of motion of the car (in which there are only the weight and the normal force), we obtain:
Substituting this expression for 
and solving for 
, we get:
Finally, plugging in the given values for the coefficient of friction and the radius of the track, we have:
It means that in its maximum value, the speed of the car is equal to 18.6m/s.
