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lys-0071
16 days ago
13

A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano

ther 4.0 km. How far west are they of their initial position? What is the magnitude of their displacement? Knight, Randall D., (Professor Emeritus). College Physics (p. 100). Pearson Education. Kindle Edition.

Physics
1 answer:
serg [2.5K]16 days ago
7 0

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

AD=AB+BD

Insert the values into the equation

D= 4+4.0\cos40^{\circ}

D=7.06\ km

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

AC=\sqrt{(CD)^2+(AD)^2}

Insert the values into the equation

AC=\sqrt{(4.0\sin40)^2+(7.06)^2}

AC=7.51

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.

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1 month ago
The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio
Yuliya22 [2420]

Answer:

the temperature on the left side is 1.48 times greater than that on the right

Explanation:

GIVEN DATA:

\gamma = 5/3

T1 = 525 K

T2 = 275 K

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P_2 = \frac{nrT_2}{v}

n and v are constant on both sides. Therefore we have

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P_1 = \frac{21}{11} P_2..............1

let the final pressure be P and the temperature T_1 {f} and T_2 {f}

P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}

P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}..................2

similarly

P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}.............3

divide equation (2) by equation (3)

\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}

T_1 {f} = 1.48 T_2 {f}

thus, the left side temperature equals 1.48 times the right side temperature

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12 days ago
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Answer:

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23 days ago
A speaker fixed to a moving platform moves toward a wall, emitting a steady sound with a frequency of 245 Hz. A person on the pl
Keith_Richards [2256]
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Response:

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