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Cerrena [4.2K]
3 years ago
5

A vector has an x component of -27.5 units and a y component of 41.4 units. Find the magnitude and direction of this vector. mag

nitude _________ unit(s) direction _____________ º counterclockwise from the +x axis
Physics
1 answer:
evablogger [386]3 years ago
4 0

Answer:

a.) magnitude __49.7__ unit(s)

b.) direction __123.6°_  counterclockwise from the +x axis

Explanation:

Let Vector is v

x-component of Vector v = x = -27.5 units   (minus sign indicate that x-component is along the minus x-axis )

y-component of Vector v = y = 41.4 units

Magnitude of v = ?

Direction of v = ?

To find the magnitude of the vector

                                     v =\sqrt{x^{2}+y^{2}  }  

                                     v = \sqrt{-27.5^{2}+41.4^{2} }

                                     v = 49.7 units  

To find direction

                                 θ = tan⁻¹(y/x)

                                 θ = tan⁻¹(41.4/-27.5)

                                 θ = -56.4°

This Angle is in the clockwise direction with respect to -x axis.

We need to find Angle counterclockwise from the +x axis.

So,

                                 θ = 180° - 56.4°

                                 θ = 123.6°                

The given vector is in 2nd quadrant

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Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

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c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

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             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

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             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

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             y = v₀ t + ½ a t²

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             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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