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3 years ago

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A physics class is building a bottle rocket. Last week, it flew for 26.189 seconds. This week, it flew for 34.9 seconds. How muc

h longer did the bottle rocket fly this week than last week?
Show work..

1 answer:

fredd [130]3 years ago

8 0

Answer:

26.189 - 34.9

Step-by-step explanation:

26.189 - 34.9 = 8.711

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Simplify √ 144x2 <br><br> A. 8x2√ 6 <br> B. 2x2√ 2 <br> C. 8√ 6x <br> D. 12x

Stella [2.4K]

√(144*x²) = (√144)*(√x²) = 12x

Answer is D.12x.

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3 years ago

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I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

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3 years ago

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What is the answer ?

kari74 [83]

Answer:

B

Step-by-step explanation:

x is alternate exterior angles and y is 180 - x

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Can someone pls pls pls help me Evaluate the expression 6.908 – g for g = 0.173.

Serjik [45]

Answer:

6.735

Step-by-step explanation:

im smart

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3 years ago

I have a roof that measures 175 ft by 45 ft. During the hurricane 11 inches of rain fell.

Nikolay [14]

Answer:

4.1 billion

Step-by-step explanation:

1 ft = 30.48 cm

1 in = 2.54 cm

The volume of rain that fell on the roof is given by ...

V = LWH

V = (175 ft × 30.48 cm/ft)(45 ft × 30.48 cm/ft)(11 in × 2.54 cm/in)

= 175×45×11×30.48²×2.54 cm³ = 204,412,236.336 cm³

At 20 drops per cm³, this will be ...

20×204,412,236.336 ≈ 4,088,244,727 . . . . raindrops

About 4.1 billion raindrops fell on your roof.

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2 years ago