Answer: displacement of airplane is 172 km in direction 34.2 degrees East of North
Explanation:
In constructing the two displacements it is noticed that the angle between the 75 km vector and the 155 km vector is a right angle (90 degrees).
Hence if the plane starts out at A, it travels to B, 75 km away, then turns 90 degrees to the right (clockwise) and travels to C, 155 km away from B. Angle ABC is 90 degrees, hence we can use Pythagoras theorem to solve for AC
AC2 = AB2 + BC2 ; AC^2 = 752 + 1552 ; from this we get AC = 172 km (3 significant figures)
Angle BAC = Tan-1(155/75) ; giving angle BAC = 64.2 degrees
Hence AC is in a direction (64.2 - 30) = 34.2 degrees East of North
Therefore the displacement of the airplane is 172 km in a direction 34.2 degrees East of North
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
A = horizontal displacement of the humming bird = 1.2 m
B = vertical displacement of the humming bird = 1.4 m
C = net displacement of the humming bird from initial to final position = ?
In the triangle drawn , Using Pythagorean theorem
C = √(A² + B²)
inserting the values
C = √(1.2² + 1.4²)
C = √(1.44 + 1.96)
C = √(3.4)
C = 1.4 m
Hence the net displacement of hummingbird comes out to be 1.4 m
Answer:
(B) (length)/(time³)
Explanation
The equation x = ½ at² + bt³ has to be dimensionally correct. In other words the term bt³ and ½ at² must have units of change of position = length.
We solve in order to find the dimension of b:
[x]=[b]*[t]³
length=[b]*time³
[b]=length/time³
In collision type of problems since momentum is always conserved
we can say
So here along with this equation we also required one more equation for the restitution coefficient
so above two equations are required to find the velocity after collision
here the change in velocity occurs due to the contact force while they contact in each other
so this is the impulse of collision while they are in contact with each other while in collision which changes the velocity of two colliding objects
