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nlexa [21]
3 years ago
12

Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.

Physics
1 answer:
Sliva [168]3 years ago
3 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is |v| = 6.93

Explanation:

From the question we are told that

    The initial point is (x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )

    The  terminal point is  (x_2 , y_2 , z_2) = (-5 , 11, 8 )

Generally the magnitude of the vector is mathematically represented as

     |v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}

=>   |v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}

=>   |v| = 6.93

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Read 2 more answers
A uniform disk of mass M = 4.9 kg has a radius of 0.12 m and is pivoted so that it rotates freely about its axis. A string wrapp
AlekseyPX

Answer:

(A) 2.4 N-m

(B) 0.035kgm^2

(C) 315.426 rad/sec

(D) 1741.13 J

(E) 725.481 rad

Explanation:

We have given mass of the disk m = 4.9 kg

Radius r = 0.12 m, that is distance = 0.12 m

Force F = 20 N

(a) Torque is equal to product of force and distance

So torque \tau =Fr, here F is force and r is distance

So \tau =20\times 0.12=2.4Nm

(B) Moment of inertia is equal to I=\frac{1}{2}mr^2

So I=\frac{1}{2}\times 4.9\times 0.12^2=0.035kgm^2

Torque is equal to \tau =I\alpha

So angular acceleration \alpha =\frac{\tau }{I}=\frac{2.4}{0.035}=68.571rad/sec^2

(C) As the disk starts from rest

So initial angular speed \omega _{0}=0rad/sec

Time t = 4.6 sec

From first equation of motion we know that \omega =\omega _0+\alpha t

So \omega =0+68.571\times  4.6=315.426rad/sec

(D) Kinetic energy is equal to KE=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 0.035\times 315.426^2=1741.13J

(E) From second equation of motion

\Theta =\omega _0t+\frac{1}{2}\alpha t^2=0\times 4.6+\frac{1}{2}\times 68.571\times 4.6^2=725.481rad

3 0
3 years ago
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