Calculating the Force of friction
1 answer:
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Answer:
Airplane speed relative to the ground is 260 km/h and θ = 22.6º direction from north to east
Explanation:
This is a problem of vector composition, a very practical method is to decompose the vectors with respect to an xy reference system, perform the sum of each component and then with the Pythagorean theorem and trigonometry find the result.
Let's take the north direction with the Y axis and the east direction as the X axis
Vy = 240 km / h airplane
Vx = 100 Km / h wind
a) See the annex
Analytical calculation of the magnitude of the speed and direction of the aircraft
V² = Vx² + Vy²
V = √ (240² + 100²)
V = 260 km/h
Airplane speed relative to the ground is 260 km/h
Tan θ = Vy / Vx
tan θ = 100/240
θ = 22.6º
Direction from north to eastb
b) What direction should the pilot have so that the resulting northbound
Vo = 240 km/h airplane
Vox = Vo cos θ
Voy = Vo sin θ
Vx = 100 km / h wind
To travel north the speeds the x axis (East) must add zero
Vx -Vox = 0
Vx = Vox = Vo cos θ
100 = 240 cos θ
θ = cos⁻¹ (100/240)
θ = 65.7º
North to West Direction
The speed in that case would be
V² = Vx² + Vy²
To go north we must find Vy
Vy² = V² - Vx²
Vy = √( 240² - 100²)
Vy = 218.2 km / h
Answer:
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.
Explanation:
Mass of object 1 , m₁ = 300 g = 0.3 kg
Mass of object 2 , m₂ = 400 g = 0.4 kg
Initial velocity of object 1 , v₁ = 5.00i-3.20j m/s
Initial velocity of object 2 , v₂ = 3.00j m/s
Mass of composite = 0.7 kg
We need to find final velocity of composite.
Here momentum is conserved.
Initial momentum = Final momentum
Initial momentum = 0.3 x (5.00i-3.20j) + 0.4 x 3.00j = 1.5 i + 0.24 j kgm/s
Final momentum = 0.7 x v = 0.7v kgm/s
Comparing
1.5 i + 0.24 j = 0.7v
v = 2.14 i + 0.34 j
Magnitude of velocity
Direction,
Velocity is 2.17 m/s at an angle of 9.03° above X-axis.