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Tatiana [17]
3 years ago
6

You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -61.0 units. A) Assuming the x co

mponent is known to be positive, specify the magnitude of the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the −x direction. B)Specify the direction of the vector. Express your answer using three significant figures.
Physics
1 answer:
kvasek [131]3 years ago
4 0

Answer:

A) magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

B)22.5° (clockwise form -ve X axis)

Explanation:

given vector = V1 = x i - 60 j

magnitude of V1 = 90

x - component can be found out by resultant formula

90^2 = x^2 + (-60)^2

x = 67.08 = 67.1 units (3sf)

FOR THE VECTOR 80 UNITS IN -VE X DIRECTION

The X component is -80-------(1)

The Y component is 0 ---------(2)

<u>For the x- component of new added vector:</u>

(1)----------- x + 67.1 = -80

x = -147.1 = -147.1

<u>For the y- component of new added vector:</u>

<u>(</u>2)---------- y - 61  = 0

y = 61.0 (3sf)

the new added vector is  = -147.1 i + 61 j

magnitude = sqrrt(147.1^2 + 61^2) = 159.2 units

direction = arctan (61 / 147.1)

               = 22.5° (clockwise form -ve X axis)

<u />

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Answer:

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\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.08 m)^{2} + \frac{1}{2}5.00 kgv^{2}                              

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.08 m)^{2}}{5.00 kg}} = 0.45 m/s

2) When the object is 5.00 cm (0.050 m) from equilibrium, the speed of the object is:

\frac{1}{2}kx^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}mv^{2}    

\frac{1}{2}280N/m(0.10 m)^{2} = \frac{1}{2}280N/m(0.05 m)^{2} + \frac{1}{2}5.00 kgv^{2}      

v = \sqrt{\frac{280N/m(0.10 m)^{2} - 280N/m(0.05 m)^{2}}{5.00 kg}} = 0.65 m/s      

 

3) When the object is at the equilibrium position, the speed of the object is:

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v = \sqrt{\frac{280N/m(0.10 m)^{2}}{5.00 kg}} = 0.75 m/s

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