Acceleration (magnitude anyway) = (change in speed) / (time for the change) .

Change in speed = (10 - 30) = -20 m/s

Time for the change = 4.0sec

Magnitude of acceleration = -20/4 = <em>-5 m/s² </em>

3 0

3 years ago

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Force X has a magnitude of 1260 pounds, and Force Y has a magnitude of 1530 pounds. They act on a single point at an angle of 4

weeeeeb [17]

Answer:

Fe= 2579.68 P

α= 24.8°

Explanation:

Look at the attached graphic

we take the forces acting on the x-y plane and applied at the origin of coordinates

FX = 1260 P , horizontal (-x)

FY = 1530 P , forming 45° with positive x axis

x-y components FY

FYx= - 1530*cos(45)° = - 1081.87 P

FYy= - 1530*sin(45)° = - 1081.87 P

Calculation of the components of net force (Fn)

Fnx= FX + FYx

Fnx= -1260 P -1081.87 P

Fnx= -2341.87 P

Fny=FYy

Fny= -1081.87 P

Calculation of the components of equilibrant force (Fe)

the x-y components of the equilibrant force are equal in magnitude but in the opposite direction to the net force components:

Fnx= -2341.87 P, then, Fex= +2341.87 P

Fny= -1081.87 P P, then, Fex= +1081.87 P

Magnitude of the equilibrant (Fe)

$F_{n} = \sqrt{(F_{nx})^{2} +(F_{ny})^{2} }$

$F_{e} =\sqrt{(2341.87)^{2}+(1081.87)^{2} }$

Fe= 2579.68 P

Calculation of the direction of equilibrant force (α)

$\alpha =tan^{-1} (\frac{F_{ny} }{F_{nx} } )$

$\alpha =tan^{-1} (\frac{1081.87 }{2341.87} )$

α= 24.8°

Look at the attached graphic

6 0

3 years ago