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skad [1K]
2 years ago
5

A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 8 feet above the bow. The rope is hauled in a

t the rate of 3 ​ft/sec. Complete parts a. and b. A triangle is drawn between a small boat and a post on the edge of a dock with a dashed vertical side labeled 8 feet, a dashed horizontal side with an arrow pointing to the right, and a solid side that rises from left to right. The intersection of the vertical side and the side that rises from left to right is labeled Ring at edge of dock. The angle at this point is labeled theta. theta theta 8' 8' a. How fast is the boat approaching the dock when 10 ft of rope are​ out? The distance between the boat and the dock is changing at a rate of nothing ​ft/sec.

Mathematics
1 answer:
choli [55]2 years ago
4 0

Answer: a) t = 1.2 second

b) dL/dt = - 1.8ft/second

Step-by-step explanation: Please find the attached files for the solution

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A pendulum is set swinging. its first oscillation is through 30 and each succeeding oscillation is through 95% of the angle of t
Marat540 [252]

<u>Answer-</u>

<em>After 76 swings</em><em> the angle through which it swings less than 1°</em>

<u>Solution-</u>

From the question,

Angle of the first of swing = 30° and then each succeeding oscillation is through 95% of the angle of the one before it.

So the angle of the second swing = (30\times \frac{95}{100})^{\circ}

Then the angle of third swing = (30\times (\frac{95}{100})^2)^{\circ}

So, this follows a Geometric Progression.

(30,\ 30\cdot \frac{95}{100},\ 30\cdot (\frac{95}{100})^2............,0)

a = The initial term = 30

r = Common ratio = \frac{95}{100}

As we have to find the number swings when the angle swept by the pendulum is less than 1°.

So we have the nth number is the series as 1, applying the formula

T_n=ar^{n-1}

Putting the values,

\Rightarrow 1=30(\frac{95}{100})^{n-1}

\Rightarrow \frac{1}{30} =(\frac{95}{100})^{n-1}

Taking logarithm of both sides,

\Rightarrow \log \frac{1}{30} =\log (\frac{95}{100})^{n-1}

\Rightarrow \log \frac{1}{30} =(n-1)\log (\frac{95}{100})

\Rightarrow -1.5=(n-1)(-0.02)

\Rightarrow 1.5=(n-1)(0.02)

\Rightarrow n-1=\dfrac{1.5}{0.02}

\Rightarrow n-1=75

\Rightarrow n=76

Therefore, after 76 swings the angle through which it swings less than 1°

8 0
2 years ago
Which expression is equivalent to 4√6/3√2
VLD [36.1K]
Given

\frac{4\sqrt{6}}{3\sqrt{2}} = \frac{4\sqrt{2\times3}}{3\sqrt{2}}  \\  \\ = \frac{4\sqrt{2}\sqrt{3}}{3\sqrt{2}} = \frac{4}{3} \sqrt{3}
5 0
2 years ago
Read 2 more answers
At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima
Firdavs [7]
\mathbf r(t)=\langle x(t),y(t),z(t)\rangle=\langle\sin t,\cos t,t\rangle

x^2+y^2+z^2=\sin^2t+\cos^2t+t^2=65
\implies t^2=64
\implies t=\pm8
7 0
2 years ago
Enter the expression 2cos2(θ)−1 , where θ is the lowercase Greek letter theta. 2cos2(θ)−1 2 c o s 2 ( θ ) − 1 = nothing
yulyashka [42]

We have to evaluate

=2 cos 2Ф - 1

⇒cos 2Ф

A =2cos²Ф-1

 B =1 -2 sin²Ф

 C = cos²Ф - sin²Ф

Possible Answers are

1.

=2 ×(2cos²Ф-1)-1

=4cos²Ф-2-1

=4cos²Ф - 3

2.

=2×(1 -2 sin²Ф)-1

=2- 4sin²Ф-1

=1 - 4sin²Ф

3.

=2× ( cos²Ф - sin²Ф) -1

=2 cos²Ф - 2sin²Ф-1

8 0
2 years ago
A cement bridge post is 24 inches square and 15 feet 9 inches in length. If the cement weighs 145 pounds per cubic foot, how muc
lisov135 [29]

Answer:

The answer is option (C), One cement bridge post weighs 9,135 pounds

Step-by-step explanation:

Step 1: Get the total volume of a cement bridge post

The cement bridge post is in the shape of a cuboid, therefor the volume of the cement bridge post can be expressed as;

Volume of cement bridge post=Base area×Length

where;

Base area=(24^2) inches square

1 foot=12 inches

Convert 24 inches to foot=24/12=2^2=4 feet²

Length=15 feet and 9 inches

1 foot=12 inches

Convert 9 inches to foot=9/12=0.75 feet

Total length=(15+0.75)=15.75 feet

replacing;

Volume of cement bridge post=(4×15.75)=63 cubic feet

Volume of cement bridge post=63 cubic feet

Step 2: Get the total weight of the cement bridge post

Total weight of the cement bridge post=Weight per cubic foot×total volume of the cement bridge post

where;

Weight per cubic foot=145 pounds per cubic foot

Total volume of the cement bridge post=63 cubic feet

replacing;

Total weight of the cement bridge post=(145×63)=9,135 pounds

One cement bridge post weighs 9,135 pounds

8 0
2 years ago
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