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Makovka662 [10]
3 years ago
5

Three forces act on a particle that moves with a constant velocity. Two of these forces are: F1 = 2.0i + 4.0j - 2.0k and F2 = -8

.0i + 9.0j - 8.0k, all components expressed in N.
What is the magnitude of the third force?
Physics
1 answer:
FrozenT [24]3 years ago
3 0

Answer:

17 N

Explanation:

Constant velocity means acceleration is 0.

If acceleration is 0, then the net force is 0.  Which means the i components add up to 0, the j components add up to 0, and the k components add up to 0.

0 = 2.0 N − 8.0 N + Fi

Fi = 6.0 N

0 = 4.0 N + 9.0 N + Fj

Fj = -13 N

0 = -2.0 N − 8.0 N + Fk

Fk = 10. N

The magnitude of the force is:

F² = Fi² + Fj² + Fk²

F² = (6.0 N)² + (-13 N)² + (10. N)²

F = √305 N

F ≈ 17 N

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Answer:

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Explanation:

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\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

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Answer: \theta=cos^{-1}0.991=7.69^o

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The angle between these two vectors can be found by:

cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\&#10;||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}

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cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o

7 0
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