consider the motion in Y-direction
v₀ = initial velocity = 29 Sin62 = 25.6 m/s
a = acceleration = - 9.8 m/s²
t = time of travel
Y = vertical displacement = - 0.89 m
using the equation
Y = v₀ t + (0.5) a t²
- 0.89 = (25.6) t + (0.5) (- 9.8) t²
t = 5.3 sec
consider the motion along the horizontal direction :
v₀ = initial velocity = 29 Cos62 = 13.6 m/s
a = acceleration = 0 m/s²
t = time of travel = 5.3 sec
X = horizontal displacement =?
using the equation
X = v₀ t + (0.5) a t²
X = (13.6) (5.3) + (0.5) (0) t²
X = 72.1 m
d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m
t = time taken = 5.3 sec
v = speed of center fielder
using the equation
v = d/t
v = 34.9/5.3
v = 6.6 m/s
<span>g = GMe/Re^2, where Re = Radius of earth (6360km), G = 6.67x10^-11 Nm^2/kg^2, and Me = Mass of earth. On the earth's surface, g = 9.81 m/s^2, so the radius of your orbit is:
R = Re * sqrt (9.81 m/s^2 / 9.00 m/s^2) = 6640km
here, the speed of the satellite is:
v = sqrt(R*9.00m/s^2) = 7730 m/s
the time it would take the satellite to complete one full rotation is:
T = 2*pi*R/v = 5397 s * 1h/3600s = 1.50 h
Hope it help i know it's long and may be confusing but if you have any more questions regarding this topic just hmu! :)</span>
Good morning.
We have:

Where
j is the unitary vector in the direction of the
y-axis.
We have that

We add the vector
-a to both sides:

Therefore, the magnitude of
b is
47 units.
Part a:
= 56
= 60
= 63
The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The
is the overall medium,
is the medium of the first half, and
is the medium of the second half.
-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.
[] See attached
Part b:
The range is 7.
The interquartile range is the range of numbers between
and
. In other words, it is 50% of the data, directly in the middle.
This becomes 63 - 56 = 7
Part c:
79 is an outlier.
It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.
-> 63 + (7 +
) ≤ 79
-> 63 + 10.5 ≤ 79
-> 73.5 ≤ 79
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.
- Heather