The answer:
the full question is as follow:
<span>A Texas rancher wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in Figure below , where A = 4.90 km and θC = 15°. He then correctly calculates the length and orientation of the fourth side D. What is the magnitude and direction of vector D?
As shown in the figure,
A + B + C + D = 0, so to find the </span>magnitude and direction of vector D, we should follow the following method:
D = 0 - (A + B + C) ,
let W = - (A + B + C), so the magnitude and direction of vector D is the same of the vector W characteristics
Magnitude
A + B + C = <span> (4.90cos7.5 - 2.48sin16 - 3.02cos15)I</span>
<span>+ (-4.9sin7.5 + 2.48cos16 + 3.02sin15)J
</span>= 1.25I +2.53J
the magnitude of W= abs value of (A + B + C) = sqrt (1.25² + 2.53²)
= 2.82
the direction of D can be found by using Dx and Dy value
we know that tan<span>θo = Dx / Dy = 1.25 / 2.53 =0.49
</span>tanθo =0.49 it implies θo = arctan 0.49 = 26.02°
direction is 26.02°
Answer:
The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .
Explanation:
Let us consider east as positive direction and west as negative direction .
Given
mass of puck 1 , 
mass of puck 2 , 
initial speed of puck 1 , 
initial speed of puck 2 , 
Final speed of puck 1 and puck 2 be
respectively
Apply conservation of linear momentum

=>
=>
-----(A)
Since collision is perfectly elastic , coefficient restitution e=1

=>
------(B)
From equation (A) and (B)

and 
Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .
Answer:
a) The velocity of rock at 1 second, v = 9.8 m/s
b) The velocity of rock at 3 second, v = 29.4 m/s
c) The velocity of rock at 5.5 second, v = 53.9 m/s
Explanation:
Given data,
The rock is dropped from a bridge.
The initial velocity of the rock, u = 0
a) The velocity of rock at 1 second,
Using the first equation of motion
v = u + gt
v = 0 + 9.8 x 1
v = 9.8 m/s
b) The velocity of rock at 3 second,
v = u + gt
v = 0 + 9.8 x 3
v = 29.4 m/s
c) The velocity of rock at 5.5 second,
v = u + gt
v = 0 + 9.8 x 5.5
v = 53.9 m/s
Answer:
Average velocity = 18 m/s
Explanation:
Given the following data;
Initial velocity = 10m/s
Acceleration = 2m/s²
Time = 4 seconds
To find the average velocity, we would use the first equation of motion;
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
V = u + at
V = 10 + 2*4
V = 10 + 8
V = 18 m/s