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3 years ago

A beetle crawls from x=1.09m to x=2.55m in 12.8s. what was the average velocity?

Physics

1 answer:

Ivan3 years ago

6 0

The average velocity of the beetle is equal to the ratio between the displacement and the time:

$v=\frac{S}{t}$

The displacement of the beetle is

$S=2.55 m-1.09 m=1.46 m$

The time is t=12.8 s

Therefore, the average velocity is

$v=\frac{S}{t}=\frac{1.46 m}{12.8 s}=0.11 m/s$

in the positive x-direction.

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Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to

Debora [2.8K]

Answer:

$\mu_{k} \approx 0.719$

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

$\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0$

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

$m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a$

$g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a$

$\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}$

$\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )$

$\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)$

$\mu_{k} \approx 0.719$

5 0

3 years ago

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:

Weight of the passenger (downwards), $m\, g$ , and possibly
Normal force $F_\text{normal}$ that the Ferris wheel exerts on this passenger (upwards.)

This passenger would feel "weightless" if the normal force on them is $0$ - that is, $F_\text{normal} = 0$ .

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:

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

3 0

1 year ago

a man can row about 4kmperhr in a still water.he rows the boat 2km up the stream and 2km back to his starting point in 2hr.how f

Nat2105 [25]

<u>Answer</u>:

The stream flowing at a speed of $2.828 \mathrm{km} / \mathrm{hr}$

<u>Explanation</u>:

Given:

Distance = 2km (both in upstream and downstream)

The speed in still water be x km/hr.

The speed in upstream = 4-x

Speed in downstream = 4+x

Solution:

We know that, Speed = distance/time

So, Time = distance/speed

Therefore,

$2=\left(\frac{2}{4-x}\right)+\left(\frac{2}{4+x}\right)$

$2=\frac{2(4+x)+2(4-x)}{(4-x)(4+x)}$

$2(4-x)(4+x)=2(4+x)+2(4-x)$

$2(4-x)(4+x)=2(4+x+4-x)$

By cancelling 2 on both sides,

$16-x^{2}=8$

$x^{2}=16-8=8$

$x=\sqrt{8}$

$x=2.828 \mathrm{km} / \mathrm{hr}$

Result:

Thus the speed of the stream is $2.828 \mathrm{km} / \mathrm{hr}$

7 0

3 years ago

Which one of the following does not accurately describe the universal gravitational law? A. The object with the smaller mass doe

ivann1987 [24]

I think the answer would be C. G<span>ravitational force is proportional to the square of the distance between the centers of both bodies. </span>

5 0

3 years ago

Read 2 more answers

A soccer ball is kicked from the top of one building with a height of H1 = 30.2 m to another building with a height of H2 = 12.0

salantis [7]

The amount of energy lost to air friction, given the data is 37.71 J

<h3>How to obtain the initial energy</h3>

Initial velocity (u) = 15.1 m/s
Mass (m) = 450 g = 450 / 1000 = 0.45 Kg
Initial Energy (E₁) = ?

E₁ = ½mu²

E₁ = ½ × 0.45 × 15.1²

E₁ = 51.3 J

<h3>How to obtain the final energy</h3>

Final velocity (u) = 19.89 m/s
Mass (m) = 450 g = 450 / 1000 = 0.45 Kg
Final Energy (E₂) = ?

E₂ = ½mv²

E₂ = ½ × 0.45 × 19.89²

E₂ = 89.01 J

<h3>How to determine the energy lost</h3>

Initial Energy (E₁) = 51.3 J
Final Energy (E₂) = 89.01 J
Energy lost =?

Energy lost = E₂ - E₁

Energy lost = 89.01 - 51.3

Energy lost = 37.71 J

Learn more about energy:

brainly.com/question/10703928

#SPJ1

8 0

1 year ago