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2 years ago

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The function f(theta) and g(theta) are sine functions where f(0) = g(0) = 0. The amplitude of f(theta) is twice the amplitude of

g(theta). The period of f(theta) is one-half the period of g(theta). If g(theta) has a period of 2x, and f(pi/4) = 4, write the function rule for g(theta). Explain your reasoning.

1 answer:

lana66690 [7]2 years ago

6 0

Because the values of both functions is 0 at ∅ = 0, both have an equilibrium position of 0.
Next, we can use the given value of f(∅) to find the amplitude:
4 = Asin(π/4)
4 = A(√2 / 2)
A = 8 / √2
We halve this amplitude to find the amplitude of g(∅):
A = 4 / √2
The period is 2π/2 = π

g(∅) = 4sin(∅/2) / √2

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Which of the following are dimensionally consistent? (Choose all that apply.)(a) a=v / t+xv2 / 2(b) x=3vt(c) xa2=x2v / t4(d) x=v

Bumek [7]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

is dimensionally consistent

B

is not dimensionally consistent

C

is dimensionally consistent

D

is not dimensionally consistent

E

is not dimensionally consistent

F

is dimensionally consistent

G

is dimensionally consistent

H

is not dimensionally consistent

Step-by-step explanation:

From the question we are told that

The equation are

$A) \ \ a^3 = \frac{x^2 v}{t^5}$

$B) \ \ x = t$

$C \ \ \ v = \frac{x^2}{at^3}$

$D \ \ \ xa^2 = \frac{x^2v}{t^4}$

$E \ \ \ x = vt+ \frac{vt^2}{2}$

$F \ \ \ x = 3vt$

$G \ \ \ v = 5at$

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$

Generally in dimension

x - length is represented as L

t - time is represented as T

m = mass is represented as M

Considering A

$a^3 = (\frac{L}{T^2} )^3 = L^3\cdot T^{-6}$

and $\frac{x^2v}{t^5 } = \frac{L^2 L T^{-1}}{T^5} = L^3 \cdot T^{-6}$

Hence

$a^3 = \frac{x^2 v}{t^5}$ is dimensionally consistent

Considering B

$x = L$

and

$t = T$

Hence

$x = t$ is not dimensionally consistent

Considering C

$v = LT^{-1}$

and

$\frac{x^2 }{at^3} = \frac{L^2}{LT^{-2} T^{3}} = LT^{-1}$

Hence

$v = \frac{x^2}{at^3}$ is dimensionally consistent

Considering D

$xa^2 = L(LT^{-2})^2 = L^3T^{-4}$

and

$\frac{x^2v}{t^4} = \frac{L^2(LT^{-1})}{ T^5} = L^3 T^{-5}$

Hence

$xa^2 = \frac{x^2v}{t^4}$ is not dimensionally consistent

Considering E

$x = L$

;

$vt = LT^{-1} T = L$

and

$\frac{vt^2}{2} = LT^{-1}T^{2} = LT$

Hence

$E \ \ \ x = vt+ \frac{vt^2}{2}$ is not dimensionally consistent

Considering F

$x = L$

and

$3vt = LT^{-1}T = L$ Note in dimensional analysis numbers are

not considered

Hence

$F \ \ \ x = 3vt$ is dimensionally consistent

Considering G

$v = LT^{-1}$

and

$at = LT^{-2}T = LT^{-1}$

Hence

$G \ \ \ v = 5at$ is dimensionally consistent

Considering H

$a = LT^{-2}$

,

$\frac{v}{t} = \frac{LT^{-1}}{T} = LT^{-2}$

and

$\frac{xv^2}{2} = L(LT^{-1})^2 = L^3T^{-2}$

Hence

$H \ \ \ a = \frac{v}{t} + \frac{xv^2}{2}$ is not dimensionally consistent

8 0

2 years ago

Se the divergence theorem to calculate the surface integral s f · ds; that is, calculate the flux of f across s. f(x, y, z) = x2

zalisa [80]

$f(x,y,z)=x^2y\,\vec\imath+xy^2\,\vec\jmath+5xyz\,\vec k$

$\implies\nabla\cdot f=2xy+2xy+5xy=9xy$

The plane has intercepts in the $x,y$ plane at (4, 0, 0) and (0, 1, 0), so the flux is given by (via the divergence theorem)

$\displaystyle\iint_Sf\cdot\mathrm dS=9\int_{x=0}^{x=4}\int_{y=0}^{y=1}\int_{z=0}^{z=4-x-4y}xy\,\mathrm dz\,\mathrm dy\,\mathrm dx=-48$

7 0

2 years ago

Find the coefficient of x2 in (3x2 – 5) (4 + 4x2), expand using algebraic expressions and answer

Scilla [17]

Answer:

- 8

Step-by-step explanation:

Given

(3x² - 5)(4 + 4x²)

Each term in the second factor is multiplied by each term in the first factor, that is

3x²(4 + 4x²) - 5 (4 + 4x²) ← distribute both parenthesis

= 12x² + 12 $x^{4}$ - 20 - 20x² ← collect like terms

= 12 $x^{4}$ - 8x² - 20

The coefficient of the x² term is - 8

7 0

2 years ago

On the planet kudzu, the probability that an animal is blue is 0.20. The probability that the animal has two heads is 0.20. Whic

Leya [2.2K]

Answer:

A.

Step-by-step explanation:

They are dependent events, indicated by the word 'given'.

7 0

2 years ago

Read 2 more answers

Wesley and brian have a total of 87 baseball cards. Wesley have 30 less than twice as many cards as brian. How many baseball car

d1i1m1o1n [39]

Brian has 39 cards and Wesley has 48 cards

Step-by-step explanation:

Let w be the number of cards Wesley has

and

b be the number of cards Brian has

Then

$w+b = 87\ \ \ \ Eqn\ 1\\w = 2b-30\ \ \ Eqn\ 2$

Putting w = 2b-30 in equation 1

$2b-30+b = 87\\3b = 87+30\\3b = 117$

Dividing both sides by 3

$\frac{3b}{3} = \frac{117}{3}\\b = 39$

Putting b=39 in equation 2

$w = 2(39)-30\\w = 78-30\\w = 48$

Hence,

Brian has 39 cards and Wesley has 48 cards

Keywords: Linear equations, variables

Learn more about linear equations at:

#LearnwithBrainly

6 0

2 years ago