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3 years ago

MathPhys Pls see this Thank you in Advance MathPhys Is the best

Physics

2 answers:

storchak [24]3 years ago

8 0

Answer:

63.8

Explanation:

umka2103 [35]3 years ago

4 0

Answer:

70 N

21°

1.1 m/s²

Explanation:

Draw a free body diagram of the block. There are three forces:

Weight pulling straight down

Normal force pushing perpendicular to the incline

Friction force pushing parallel to the incline

Part 1

Sum the forces in the perpendicular direction:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

The block is at rest, so F = N μs:

F = N μs

F = mg μs cos θ

F = (20 kg) (9.8 m/s²) (0.38) (cos 19°)

F = 70 N

Part 2

Sum the forces in the parallel direction (down the incline is positive):

∑F = ma

mg sin θ − F = 0

mg sin θ = N μs

mg sin θ = mg μs cos θ

tan θ = μs

θ = atan μs

θ = atan 0.38

θ = 21°

Part 3

Sum the forces in the parallel direction (this time, acceleration is not 0).

∑F = ma

mg sin θ − F = ma

mg sin θ − N μk = ma

mg sin θ − mg μk cos θ = ma

a = g (sin θ − μk cos θ)

a = (9.8 m/s²) (sin 24° − 0.32 cos 24°)

a = 1.1 m/s²

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Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

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How many revolutions per minute would a 23 m -diameter Ferris wheel need to make for the passengers to feel "weightless" at the

kirza4 [7]

Answer:

Approximately $6.2\; {\rm rpm}$ , assuming that the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ .

Explanation:

Let $\omega$ denote the required angular velocity of this Ferris wheel. Let $m$ denote the mass of a particular passenger on this Ferris wheel.

At the topmost point of the Ferris wheel, there would be at most two forces acting on this passenger:

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The net force on this passenger is $(m\, g - F_\text{normal})$ . Hence, when $F_\text{normal} = 0$ , the net force on this passenger would be equal to $m\, g$ .

Passengers on this Ferris wheel are in a centripetal motion of angular velocity $\omega$ around a circle of radius $r$ . Thus, the centripetal acceleration of these passengers would be $a = \omega^{2}\, r$ . The net force on a passenger of mass $m$ would be $m\, a = m\, \omega^{2}\, r$ .

Notice that $m\, \omega^{2} \, r = (\text{Net Force}) = m\, g$ . Solve this equation for $\omega$ , the angular speed of this Ferris wheel. Since $g = 9.81\; {\rm m\cdot s^{-2}}$ and $r = 23\; {\rm m}$ :

$\begin{aligned} \omega^{2} = \frac{g}{r}\end{aligned}$ .

$\begin{aligned} \omega &= \sqrt{\frac{g}{r}} \\ &= \sqrt{\frac{9.81\; {\rm m \cdot s^{-2}}}{23\; {\rm m}}} \\ &\approx 0.653\; {\rm rad \cdot s^{-1}} \end{aligned}$ .

The question is asking for the angular velocity of this Ferris wheel in the unit ${\rm rpm}$ , where $1\; {\rm rpm} = (2\, \pi\; {\rm rad}) / (60\; {\rm s})$ . Apply unit conversion:

$\begin{aligned} \omega &\approx 0.653\; {\rm rad \cdot s^{-1}} \\ &= 0.653\; {\rm rad \cdot s^{-1}} \times \frac{1\; {\rm rpm}}{(2\, \pi\; {\rm rad}) / (60\; {\rm s})} \\ &= 0.653\; {\rm rad \cdot s^{-1} \times \frac{60\; {\rm s}}{2\, \pi\; {\rm rad}} \times 1\; {\rm rpm} \\ &\approx 6.2\; {\rm rpm} \end{aligned}$ .

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