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swat32
3 years ago
11

A 1550 kg car located at < 319, 0, 0 > m has a momentum of < 45000, 0, 0 > kg·m/s. What is its location 10 s later?

(Express your answer in vector form.)
Physics
1 answer:
Serggg [28]3 years ago
5 0

Answer:

< 45319, 0, 0 > m

Explanation:

mass, m = 1550 kg

r = < 319, 0, 0 > m

v = < 4500, 0, 0 > m/s

t = 10 s

distance traveled in 10 second is

d =  v x t

d = < 4500 x 10, 0 x 10, 0 x 10 > m

d = < 45000, 0, 0 > m

So, the new location of the car is

r' = r + d = < 319, 0, 0 > + < 45000, 0, 0 >

r' = < 45319, 0, 0 > m

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The I3 will be 158 A.

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To calculate I3 firstly, V4 has to be calculated,

V_{4} =I_{4} R_{4}

V_{4} = V_{2} / R_{4} + R_{5}  * R_{4}

V_{4} = 12 * 135 / 135+61

V_{4} = 8.26V

For I3,

I_{3} = R_{1} /(R1+R3 + (R1+R3)(R2+R6) * (V2 - V1 (R1+R2+R6/R1)

I3=(61)/((61)(50)+(61+50)(141+141)) (12 -18 (1+(141+141)/61)) = -.158 A

Hence, the current through I3 will be 158 A.

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Answer:

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The acceleration due to gravity is given by

g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

a_c=\omega^2R

a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s

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