3.8q + 21.2 + 8.3r
The perimeter is all sides added together, therefore our equation looks like this:
However, we can get rid of the parentheses and add like terms together.
A small plane can carry a total of 1 Ton.
2 answers:
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Answer:
a) 9.434 m/s
b) i (2+5*t) + (1+8*t-4.905*t²) j
c) t= 8/5 secs
d) 3.598 m/s
e) See explanation
Step-by-step explanation:
Part a)
The speed of the ball can be calculated from the given velocity v = 5i +8j
Taking magnitude of v = 5i + 8j
magnitude (v) = = 9.434 m/s.
Part b)
Using kinematic equation of particle as follows:
Sf = Si + Vi*t + 0.5*a*t² ..... Eq 1
Given: Si = (2i + j) m ; Vi = (5i+8j) m/s; a = -9.81 j m/s²
We evaluate Eq 1:
Sf = (2i+j) + (5i+8j)*t + 0.5*(-9.81j)*t²
We get after combining similar terms:
Sf = i (2+5*t) + (1+8*t-4.905*t²) j ..... Eq 2
Part c)
Using kinematic equation of particle only in i axis as follows we use Eq 1:
Sf = Si + Vi*t + 0.5*a*t²
Given: Si = 2 m ; Sf = 10; Vi = 5 m/s; a = 0;
We evaluate Eq 1:
10 = 2 + 5*t - Solve for t
t = 8/5 seconds
Note: The above is the time t when the ball is due north of (10i+7j) i.e having a position vector of 10 in east direction but unknown in north direction. A point directly above or below 10i + 7j.
Part d)
The interception of ball and the player occurs at the same t = 8/5 secs and @ position vector (10i + aj) where a is a constant needs to be found.
Find a:
Using Eq 2 found in part b:
Sf = i (2+5*t) + (1+8*t-4.905*t²) j
Evaluate @ t= 8/5 secs
Sf = (10) i + (1.2432) j .... Eq 3
To find the speed v of the player when he intercepts the ball at Sf = (10) i + (1.2432) j is evaluated as follows:
v = change in position of player / Time
Hence, v = -3.598 j = 3.598 m/s
Part e)
Friction between the ball and surface from which is launched.