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2 years ago

Vector u has a magnitude of 5sq rt 3 feet and is directed due west. Vector v has a magnitude

Mathematics

1 answer:

Vikki [24]2 years ago

7 0

The magnitude and the direction angle for the resultant vector are 10N and 120degrees respectively

The formula for calculating the resultant vector is expressed as:

$R^2 = u^2 + v^2\\R =\sqrt{u^2+v^2}$

Given the following distances:

u = -5√3 feet (west direction)

v = 5 feet

Substitute the given values into the formula above;

$R=\sqrt{(5\sqrt{3} )^2+5^2} \\R=\sqrt{(25 \times 3)+25}\\R=\sqrt{75 + 25}\\R=\sqrt{100}\\R = 10 feet$

The magnitude of the resultant vector is 10feet

Get the direction:

$\theta = tan^{-1}\frac{y}{x} \\\theta = tan^{-1}\frac{5}{-5\sqrt{3} }\\\theta =tan^{-1}\frac{-1}{\sqrt{3} }\\\theta = -60^0$

Since tan is negative in the second quadrant, hence;

$\theta = 180 - 60\\\theta = 120^0$

Hence the magnitude and the direction angle for the resultant vector are 10N and 120degrees respectively.

Learn more here: brainly.com/question/5177683

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in a church, the ratio of adults to children is 6 to 11. if there are 306 people in the church, how many children are there?

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198

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3 years ago

Select the curve generated by the parametric equations. Indicate with an arrow the direction in which the curve is traced as t i

bixtya [17]

Answer:

length of the curve = 8

Step-by-step explanation:

Given parametric equations are x = t + sin(t) and y = cos(t) and given interval is

−π ≤ t ≤ π

Given data the arrow the direction in which the curve is traces means

the length of the curve of the given parametric equations.

The formula of length of the curve is

$\int\limits^a_b {\sqrt{\frac{(dx}{dt}) ^{2}+(\frac{dy}{dt}) ^2 } } \, dx$

Given limits values are −π ≤ t ≤ π

x = t + sin(t) ...….. (1)

y = cos(t).......(2)

differentiating equation (1) with respective to 'x'

$\frac{dx}{dt} = 1+cost$

differentiating equation (2) with respective to 'y'

$\frac{dy}{dt} = -sint$

The length of curve is

$\int\limits^\pi_\pi {\sqrt{(1+cost)^{2}+(-sint)^2 } } \, dt$

$\int\limits^\pi_\pi \, {\sqrt{(1+cost)^{2}+2cost+(sint)^2 } } \, dt$

on simplification , we get

here using sin^2(t) +cos^2(t) =1 and after simplification , we get

$\int\limits^\pi_\pi \, {\sqrt{(2+2cost } } \, dt$

$\sqrt{2} \int\limits^\pi_\pi \, {\sqrt{(1+1cost } } \, dt$

again using formula, 1+cost = 2cos^2(t/2)

$\sqrt{2} \int\limits^\pi _\pi {\sqrt{2cos^2\frac{t}{2} } } \, dt$

Taking common $\sqrt{2}$ we get ,

$\sqrt{2}\sqrt{2} \int\limits^\pi _\pi ( {\sqrt{cos^2\frac{t}{2} } } \, dt$

$2(\int\limits^\pi _\pi {cos\frac{t}{2} } \, dt$

$2(\frac{sin(\frac{t}{2} }{\frac{t}{2} } )^{\pi } _{-\pi }$

length of curve = $4(sin(\frac{\pi }{2} )- sin(\frac{-\pi }{2} ))$

length of the curve is = 4(1+1) = 8

conclusion:-

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3 years ago

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3 years ago