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solong [7]
4 years ago
14

A hiker begins a trip by first walking 25.0 km southeast from her car. She stops and sets up her tent for the night. On the seco

nd day, she walks 40.0 km in a direction 60.0° north of east, at which point she discovers a forest ranger’s tower. (a) Determine the components of the hiker’s displacement for each day. (b) Determine the magnitude and direction of the total displacement.
Physics
1 answer:
Sunny_sXe [5.5K]4 years ago
6 0

Answer:

(a)

x_1=17.68km\\y_1=-17.68km\\x_2=20km\\y_2=34.64km

(b) r=41.32km

\alpha =24.23^o

Explanation:

Let us take the north direction to be the positive y-axis and the east to be positive x-axis.

First day:

25.0 km southeast, which implies 45^o south of east. The y-component will be negative and the x-component will be positive.

x_1=25cos45^o=17.68km

y_1=-25sin45^o=-17.68km

Second day:

She starts off at the stopping point of last day. This time, both the y- and x-components are positive.

x_2=40cos60^o=20km

y_2=40sin60^o=34.64km

Therefore, total displacements:

x=x_1+x_2=(17.68+20)km=37.68km

y=y_1+y_2=(-17.68+34.64)km=16.96km

Magnitude of displacements,

r=\sqrt{x^2+y^2}=41.32km

Direction,

\alpha =tan^{-1}(\frac{y}{x})=24.23^o

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The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

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the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

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6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

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