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enyata [817]
2 years ago
7

A boy is pulling a 150 kg crate with a force of 1150 N. If the crate experiences a frictional force of 490 N, find:

Physics
1 answer:
valentina_108 [34]2 years ago
7 0
<h3><u>For the aceleration:</u></h3>

First, let's find the resultant, and <u>applicate 2nd law of Newton</u> using the resultant, so:

R = ma

F - Ff = ma

Data:

F = Force = 1150 N

Ff = Friction force = 490 N

m = Mass = 150 kg

a = Aceleraction = ?

Replacing according our data:

1150 N - 490 N = 150 kg * a

660 N = 150 kg * a

660 N / 150 kg = a

a = 4,4 m/s²   ← Aceleration of the object

<h3><u>For the normal force:</u></h3>

The normal force IS NOT the resultant force, the normal force's the force between the ground and the object, in another words, is the weight of the object, and for the weight:

w = mg

w = 150 kg * 10 m/s²

w = 1500 N   ← Normal force between object and ground.

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Find the cube roots of 27(cos 327° + i sin 327° ). Write the answer in trigonometric form.
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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

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