What does the area under a curve of a graph of velocity versus time yield? A. Average acceleration O B. Average velocity OC.Tota
2 answers:
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Answer:
A = -14.87 i ^ + 8.42 j ^ + 0 k ^
B = -25.41 i ^ -12.0 j ^ + 0 k ^
Explanation:
For this exercise let's use trigonometry by decomposing to vectors
vector A
module 17.1 with an angle of 150.5 counterclockwise.
Sin 150.5 = / A
cos 150.5 = Ax / A
A_{y} = A sin 150.5 = 17.1 sin 150.5
Aₓ = A cos 1505 = 172 cos 150.5
A_{y} = 8,420
Aₓ = -14.870
the vector is
A = -14.87 i ^ + 8.42 j ^ + 0 k ^
Vector B
= 28.1 sin 205.3
Bₓ = 28.1 cos 205.3
B_{y} = -12.009
Bₓ = -25.405
the vector is
B = -25.41 i ^ -12.0 j ^ + 0 k ^
Answer:
U = - 4 x
Explanation:
ΔV = potential difference = Volts
q = charge on electron = C
electric potential energy is given as
U = q ΔV = ( ) (
)
= - 4 x