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Vlad [161]
3 years ago
14

A horse trots 17 miles north and then turns and trots 24 miles at an angle of 40° west of north. Find the resultant displacement

and direction mathematically. Hint: This is a non-perpendicular vector problem.
Physics
1 answer:
DanielleElmas [232]3 years ago
8 0

Answer:

 Resultant displacement = 38.58 miles at an angle of 23.57° north of west.

Explanation:

Let north represent positive y direction and east represent positive x direction. Unit vector along x direction is represented by i and vector along unit vector along y direction is represented by j

Initial displacement = 17 miles north = 17 j miles

Final displacement = 24 miles at an angle of 40° west of north = 24 miles at an angle of 130° to positive x -axis

                               = (24 cos 130 i + 24 sin 130 j )

                               = (-15.43 i + 18.36 j ) miles

Total displacement = 17 j + (-15.43 i + 18.36 j) = -15.43 i + 35.36 j

Magnitude of displacement =\sqrt{(-15.43)^2+35.36^2} =38.58 miles

Angle to positive x -axis = tan⁻¹(35.36/-15.43) = 113.57°

Resultant displacement = 38.58 miles at an angle of 23.57° west of north.

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Explanation:

Period=1/frequency

Period=1/295 seconds

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4 years ago
Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.
Sati [7]

Answer:

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836, z^{\frac{1}{3} }= -1.967 - i\cdot 2.265, z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

Explanation:

The cube root of the complex number can determined by the following De Moivre's Formula:

z^{\frac{1}{n} } = r^{\frac{1}{n} }\cdot \left[\cos\left(\frac{x + 2\pi\cdot k}{n} \right) + i\cdot \sin\left(\frac{x+2\pi\cdot k}{n} \right)\right]

Where angles are measured in radians and k represents an integer between 0 and n - 1.

The magnitude of the complex number is 27 and the equivalent angular value is 1.817\pi. The set of cubic roots are, respectively:

k = 0

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{1.817\pi}{3} \right)+i\cdot \sin\left(\frac{1.817\pi}{3} \right)]

z^{\frac{1}{3} }= -0.978 + i\cdot 2.836

k = 1

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{3.817\pi}{3} \right)+i\cdot \sin\left(\frac{3.817\pi}{3} \right)]

z^{\frac{1}{3} }= -1.967 - i\cdot 2.265

k = 2

z^{\frac{1}{3} } = 3\cdot \left[\cos \left(\frac{5.817\pi}{3} \right)+i\cdot \sin\left(\frac{5.817\pi}{3} \right)]

z^{\frac{1}{3} }= 2.945 - i\cdot 0.571

5 0
3 years ago
A bag of sugar weighs 3.50 lb on Earth. What would it weigh in newtons on the Moon, where the free-fall acceleration isone-sixth
Alex73 [517]

Answer:

F_{Earth}= 15.57 N

F_{Moon}= 2.60 N

F_{Uranus}= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

Explanation:

The weight of the sugar bag on Earth is:

g=9.81 m/s²

m=3.50 lb=1.59 kg

F_{Earth}=m·g=1.59 kg×9.81 m/s²= 15.57 N

The weight of the sugar bag on the Moon is:

g=9.81 m/s²÷6= 1.635 m/s²

F_{Moon}=m·g=1.59 kg× 1.635 m/s²= 2.60 N

The weight of the sugar bag on the Uranus is:

g=9.81 m/s²×1.09=10.69 m/s²

F_{Uranus}=m·g=1.59 kg×10.69 m/s²= 16.98 N

The mass of the bag is the same on the three planets. m=1.59 kg

5 0
3 years ago
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