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3 years ago

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Consider three vectors A⃗ ,B⃗ and C⃗ all lying in the xy plane. Vector A⃗ has length 6.2 and makes an angle of 131 degrees with

the x-axis. Vector B⃗ has length 7.3 and makes an angle of 32 degrees with the x-axis. The vector D⃗ =A⃗ +B⃗ +C⃗ has x-component DX = 3.1 and y-component DY = 6.7 .A) Find the magnitude of the vector C⃗ .B) Find the angle in degrees that vector C⃗ makes measured counterclockwise from the x-axis.

1 answer:

vagabundo [1.1K]3 years ago

8 0

Answer:

A) 2.09

B) -62.09°

Explanation:

If D = A+B+C then C = D - A - B

Converting both A and B to rectangular coordinates, we get:

A = [-4.07 , 4.68] and B = [6.19 , 3.87]

Replacing these values we can get the x and y components of C:

C = [0.98 , -1.85]

Its module is:

$|C| = \sqrt{0.98^2+(-1.85)^2}=2.09$

Its angle is:

$\alpha _c = atan(-1.85/0.98) = -62.09\°$

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r=0.127

Explanation:

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10 I (R₁R₂/(R₂+ R₁)) = I (R₁+R₂)

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Divides by R₂²

8R₁/R₂ = ( R₁/R₂)²+ 1

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Explanation:

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$F_{net} = -mg Sin (\theta)$

or, ma = -mg Sin (\theta)[/tex]

a = $-g Sin (\theta)$

= $-9.8 \times Sin (20^{o})$

= -3.35 $m/s^{2}$

According to the kinematic equation of motion,

$v^{2} - v^{2}_{o} = 2as$

Distance traveled by the block before stopping is as follows.

s = $\frac{v^{2} - v^{2}_{o}}{2a}$

= $\frac{(0)^{2} - (12.0)^{2}_{o}}{2 \times -3.35}$

= 21.5 m

According to the kinematic equation of motion,

v = $v_{o} + at$

0 = $12.0 m/s + \frac{1}{2} \times -3.35 m/s^{2} \times t$

$t_{1}$ = 7.16 sec

Therefore, before coming to rest the surface of the plane will slide the box till 7.16 sec.

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$F_{net} = -mg Sin (\theta)$

ma = $mg Sin (\theta)$

a = $g Sin (\theta)$

= $9.8 m/s^{2} \times Sin (20^{o})$

= 3.35 $m/s^{2}$

Kinematics equation of the motion is as follows.

s = $v_{o}t + \frac{1}{2}at^{2}$

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$t_{2}$ = $\sqrt{\frac{2 \times 21.5 m}{3.35 m/s^{2}}}$

= 3.58 sec

Hence, total time taken by the block to return to its starting position is as follows.

t = $t_{1} + t_{2}$

= 7.16 sec + 3.58 sec

= 10.7 sec

Thus, we can conclude that 10.7 sec time it take to return to its starting position.

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