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7nadin3 [17]
3 years ago
8

A light plane is headed due south with a speed of 200 km/h relative to still air. After 1.00 hour, the pilot notices that they h

ave covered only 137 km and their direction is not south but 15.0∘ east of south. What is the wind velocity?
Physics
1 answer:
Lilit [14]3 years ago
8 0

Answer: the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

Explanation: The initial velocity of the plane is 200km/h south (supose that south is our positive x-axis here and east is the positive y-axis)

In one hour, the plane is located 137km away from the initial position, and the position in X is equal to 137km*cos(15°) = 132.33, this means that the velocity in the x axis is equal to 132.33 km/h, knowing that the initial velocity of the plane was 200km in the x-axis, this means that the velocity of the air must be:

132.33km/h - 200km/h = -67.69km/h

km and the position in "y" is equal to 137km*sin(15°) = 35.4km

This means that the velocity of the air in the y-axis is 35.4km/h

So the total velocity of the air is 67.69km/h to the north and 35.4km/h to the east.

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A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north
Sedaia [141]

Answer:

Magnitude of displacement = 2.07 km

Magnitude of average velocity = 1.17 kmph

Explanation:

Let east represent positive x axis and north represent positive y axis.

A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.

1.93 km due wast

           s ₁ = 1.93 i km

1.03 km due south

           s₂ = -1.03 j km

3.84 km in a direction 52.8 ° north of west

           s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km

Total displacement

          s = s ₁+  s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j

  Magnitude of displacement, =\sqrt{(-0.39)^2+2.03^2}=2.07km

Time taken = 1.771 hour

Magnitude of average velocity, =\frac{2.07}{1.771}=1.17km/hr

7 0
3 years ago
Two moles of helium are initially at a temperature of 21.0 ∘Cand occupy a volume of 3.30×10−2 m3 . The helium first expands at c
Anettt [7]

Answer:

(B) The total internal energy of the helium is 4888.6 Joules

(C) The total work done by the helium is 2959.25 Joules

(D) The final volume of the helium is 0.066 cubic meter

Explanation:

(B) ∆U = P(V2 - V1)

From ideal gas equation, PV = nRT

T1 = 21°C = 294K, V1 = 0.033m^3, n = 2moles, V2 = 2× 0.033=0.066m^3

P = nRT ÷ V = (2×8.314×294) ÷ 0.033 = 148140.4 Pascal

∆U = 148140.4(0.066 - 0.033) = 4888.6 Joules

(C) P2 = P1(V1÷V2)^1.4 =148140.4(0.033÷0.066)^1.4= 148140.4×0.379=56134.7 Pascal

Assuming a closed system

(C) Wc = (P1V1 - P2V2) ÷ 0.4 = (148140.4×0.033 - 56134.7×0.066) ÷ 0.4 = (4888.6 - 3704.9) ÷ 0.4 = 1183.7 ÷ 0.4 = 2959.25 Joules

(C) Final volume = 2×initial volume = 2×0.033= 0.066 cubic meter

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Explanation: hope this helped

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