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Advocard [28]
2 years ago
15

consider the points ​p(​,​,​) and ​q(​,​,​). a. find and state your answer in two​ forms: and aibjck. b. find the magnitude of .

c. find two unit vectors parallel to .
Physics
1 answer:
astra-53 [7]2 years ago
6 0

Consider the points <u>p(,,) and q(,,):</u>

P (-4,1,0)

Q (-4,-5,2)

A. find and state your answer in two forms

PQ = <u>(-4+4, -5-1, 2-0)</u>

= <u>(0,-6,2)</u>

B. find the magnitude

 |PQ|= √(0)2+ (-6)2 + (2)2

= √36+4

= √40

= <u>2√10</u>

C. find two unit vectors

Unit vector parallel to PQ

= < 0, -3/√10, 1/√10 >

= <u>< 0, 3/√10, -1/√10 ></u>

In physics, magnitude is defined as the maximum size and orientation of an object. Size is used as a common factor for vector size and scalar size. By definition, we know that a scalar quantity is one that has only one quantity.

Force Magnitude is a numerical value that expresses the strength of a force. Example: Suppose the force is = 10 N towards the east. "East" is the direction, "10" is the strength.

Learn more about magnitude here:- brainly.com/question/24468862

#SPJ4

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
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Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
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