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Annette [7]
3 years ago
9

An ant is crawling along a yardstick that is pointed with the 0-inch mark to the east and the 36-inch mark to the west. It start

s at the 16 inch mark, crawls to the 29-inch mark, then moves to the 14-inch mark. What is the total distance the ant traveled? What was the total displacement of the ant?
Physics
1 answer:
FrozenT [24]3 years ago
6 0

Answer:

  • The total distance traveled is 28 inches.
  • The displacement is 2 inches to the east.

Explanation:

Lets put a frame of reference in the problem. Starting the frame of reference at the point with the 0-inch mark, and making the unit vector \hat{i} pointing in the west direction, the ant start at position

\vec{r}_0 = 16 \ inch \ \hat{i}

Then, moves to

\vec{r}_1 = 29 \ inch \ \hat{i}

so, the distance traveled here is

d_1 = |\vec{r}_1 - \vec{r}_0  | = | 29 \ inch   \ \hat{i} - 16 \ inch   \ \hat{i}  |

d_1 =  | 13 \ inch   \ \hat{i}  |

d_1 =  13 \ inch

after this, the ant travels to

\vec{r}_2 = 14 \ inch \ \hat{i}

so, the distance traveled here is

d_2 = |\vec{r}_2 - \vec{r}_1  | = | 14 \ inch   \ \hat{i} - 29 \ inch   \ \hat{i}  |

d_2 =  | - 15 \ inch   \ \hat{i}  |

d_2 =  15 \ inch

The total distance traveled will be:

d_1 + d_2 = 13 \ inch + 15 \ inch = 28 \ inch

The displacement is the final position vector minus the initial position vector:

\vec{D}=\vec{r}_2 - \vec{r}_1

\vec{D}= 14 \ inch   \ \hat{i} - 16 \ inch \ \hat{i}

\vec{D}= - 2 \ inch \ \hat{i}

This is 2 inches to the east.

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tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

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a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

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u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

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1 year ago
A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down
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Answer:

Explanation:

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We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

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R = 350ft

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a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

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a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

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PART C)

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a_r = \frac{21.5^2}{350}

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a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

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Answer:

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This can be obtained as follow:

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d is the density of the liquid.

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Hope it helps you

Good luck

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