The answer is Golgi apparatus.
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It's D! hope I helped.
consider the motion along the X-direction
X = horizontal displacement = 80 m
= initial velocity along the x-direction = v Cos60
t = time of travel
using the equation
X = t
80 = (v Cos60) (t)
t = 160/v eq-1
consider the motion in vertical direction :
Y = vertical displacement = 20 m
= initial velocity in Y-direction = v Sin60
a = acceleration = - 9.8 m/s²
t = time of travel = 160/v
using the equation
Y = t + (0.5) a t²
20 = (v Sin60) (160/v) + (0.5) (- 9.8) (160/v)²
v = 32.5 m/s
Answer:
Explanation:
a)Magnitude = 
84=
x= +50.67 or -50.67 units
b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.
To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.
Magnitude = 
= 146.85 units
c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - 
i.e 152.85 degrees from the +ve x-axis.
Answer:
For elliptical orbits: seldom
For circular orbits: always
Explanation:
We start by analzying a circular orbit.
For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.
Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.
So for a circular orbit,
the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.
Now we analyze an elliptical orbit.
An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.
Therefore, for an elliptical orbit,
the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.
