In: Physics
Part A..
Let, Given vector = R = Rx i + Ry j
here, |R| = 82.0 units
Ry = -58.0
So,
|R|^2 = Rx^2 + Ry^2
Rx^2 = 82.0^2 - (-58.0)^2 = 3360
Rx = sqrt(3360) = 57.97
Rx = 58.0 unit
Rx = -58.0, +58.0
Part-B.
Given, x-component is known to be positive.
then, R = 58.0 i - 58.0 j
Let, anothe vector is,
R' = R'x i + R'y j
given,
R + R' = -80 i + 0 j
So,
(58.0 i - 58.0 j) + (R'x i + R'y j) = -80 i + 0 j
(58.0 + R'x) i + (-58.0 + R'y) j = -80 i + 0 j
equating both side,
58.0 + R'x = -80
R'x = -80.0-58.0 = -138
-58.0 + R'y = 0
R'y = 58.0 j
then,
R' = -138 i + 58.0 j
Magnitude will be:
|R'| = sqrt(138^2 + 58.0^2)
|R'| = 150
Part-C.
direction will be:
= arctan(R'y/R'x)
= arctan(58.0/(-138))
= 22.8 deg
Part A..
Let, Given vector = R = Rx i + Ry j
here, |R| = 82.0 units
Ry = -58.0
So,
|R|^2 = Rx^2 + Ry^2
Rx^2 = 82.0^2 - (-58.0)^2 = 3360
Rx = sqrt(3360) = 57.97
Rx = 58.0 unit
Rx = +58.0, -58.0
Part-B.
Given, x-component is known to be positive.
then, R = 58.0 i - 58.0 j
Let, anothe vector is,
R' = R'x i + R'y j
given,
R + R' = -80 i + 0 j
So,
(58.0 i - 58.0 j) + (R'x i + R'y j) = -80 i + 0 j
(58.0 + R'x) i + (-58.0 + R'y) j = -80 i + 0 j
equating both side,
58.0 + R'x = -80
R'x = -80.0-58.0 = -138
-58.0 + R'y = 0
R'y = 58.0 j
then,
R' = -138 i + 58.0 j
Magnitude will be:
|R'| = sqrt(138^2 + 58.0^2)
|R'| = 150
Part-C.
direction will be:
= arctan(R'y/R'x)
= arctan(58.0/(-138))
= -22.8 deg above -x axis.
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"Let me know if you have any query."