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3 years ago

15

a particle of mass 1.3kg is sliding down a frictionless slope inclined at 30 to the horizontal. the acceleration of the particle

down the slope is

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

$4.9\ m/sec^2$

Explanation:

The computation of acceleration of the particle down the slope is shown below:-

data provided in the question

Particle of mass = 1.3 kg i,e sliding down

Inclined = 30 to the horizontal

based on the above information

Force is given by

$N = mg\ cos \theta$ ............ 1

and sliding force is given by

$F = mg\ sin\alpha$

$a = g(sin\ 30^{\circ})$

$= 9.8\times \frac{1}{2} m/sec^2$

= $= 4.9\ m/sec^2$

Hence, the acceleration of the particle down the slope is 4.9 m/sec^2

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Answer:

From Wikipedia:

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Six artificial satellites circle a space station at constant speed. The mass m of each satellite, distance L from the space stat

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a) $T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

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Explanation:

a) Since we are told the satellites circle the space station at constant speed, we can assume they follow a uniform circular motion and their tangential speeds $V$ are given by:

$V=\omega L=\frac{2\pi}{T} L$ (1)

Where:

$\omega$ is the angular frequency

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Isolating $T$ :

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Applying this equation to each satellite:

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$T_{5}=\frac{2 \pi L}{V_{5}}=785.398 s$ (7)

$T_{6}=\frac{2 \pi L}{V_{6}}=196.349 s$ (8)

Ordering this periods from largest to smallest:

$T_{2}>T_{5}>T_{1}>T_{3}=T_{6}>T_{4}$

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$a=\frac{V}{T}$ (9)

Applying this equation to each satellite:

$a_{1}=\frac{V_{1}}{T_{1}}=0.458 m/s^{2}$ (10)

$a_{2}=\frac{V_{2}}{T_{2}}=0.0254 m/s^{2}$ (11)

$a_{3}=\frac{V_{3}}{T_{3}}=0.4074 m/s^{2}$ (12)

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$a_{5}=\frac{V_{5}}{T_{5}}=0.101 m/s^{2}$ (14)

$a_{6}=\frac{V_{6}}{T_{6}}=0.814 m/s^{2}$ (15)

Ordering this acceerations from largest to smallest:

$a_{4}>a_{6}>a_{1}>a_{3}>a_{5}>a_{2}$

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4 years ago