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1 year ago

What is the x component of a vector that is defined as 45m at -35°?

Physics

1 answer:

goldenfox [79]1 year ago

4 0

the x- component of the vector is 36.86 m.

<h3>What is a vector?</h3>

Vectors are quantities that have both magnitude and direcion

To calculate the x-component of the vector, we use the formula below.

Formula:

dx = dcosθ.......... Equation 1

Where;

dx = x-component of the vector
d = vector between the x-y component
θ = Angle of the vector to the horizontal.

From the question,

Given:

d = 45 m
θ = -35°

Substitute these values into equation 1

dx = 45cos(-35°)
dx = 45×0.918
dx = 36.86 m.

Hence, the x- component of the vector is 36.86 m.

Learn more about vector here: brainly.com/question/25705666

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a) $\theta = 52\,rad$ , b) $v_{A} = 44\cdot r \,\frac{m}{s}$ , c) $a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$ , $a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

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a) The angular motion is obtained by integrating the angular acceleration function twice:

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The angular motion when t = 2 s. is:

$\theta = \frac{1}{4}\cdot (2\,s)^{4} + 6\cdot (2\,s)^{2} + 12\cdot (2\,s)$

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b) Let be $r$ the distance between A and the rotation axis, measured in meters. The magnitude of the angular velocity when t = 2 s. is:

$\omega = (2\,s)^{3} + 12\cdot (2\,s) + 12$

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$\alpha = 3\cdot (2\,s)^{2} + 12$

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$a_{A,n} = \omega^{2}\cdot r$

$a_{A,n} = \left(44\right)^{2}\cdot r$

$a_{A,n} = 1936\cdot r\,\frac{m}{s^{2}}$

$a_{A,t} = \alpha \cdot r$

$a_{A,t} = 24\cdot r\,\frac{m}{s^{2}}$

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3 years ago

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Answer:

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Explanation:

Applying,

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