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3 years ago

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A rope, attached to a weight, goes up through a pulley at the ceiling and back down to a worker. The worker holds the rope at th

e same height as the connection point between the rope and weight. The distance from the connection point to the ceiling is 40 ft. Suppose the worker stands directly next to the weight (i.e., a total rope length of 80 ft) and begins to walk away at a constant rate of 2 ft/s. How fast is the weight rising when the worker has walked: 10 feet? Answer = 30 feet? Answer =

1 answer:

Mrac [35]3 years ago

4 0

Answer:0.246ft/s

Step-by-step explanation: when the the worker starts moving and stops 10 feets away from lateral rope connecting the load, it forms a right angle triangle as see in the pix.

V = s/t ... equ 1

S= distance traveled by worker while pulling the rope = 10 ft

V = velocity of worker = 2ft/s

T = time taken the worker to move 10 ft = time taken for the load to move up = 5seconds

Use Pythagoras theorem to determine distance from pulley to new position of worker

Square root of (40^2 + 10^2) = 41.23ft

Distance traveled by load = D = 41.23 - 40

=1.23ft

Speed of load = D/ t...equ2

Substituting D and the into equ2

S.l = speed of load = 1.23/5 =0.246ft/s

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natima [27]

Answer:

4.9 feet

Step-by-step explanation:

First, we need to determine the height of the building using the following:

$L^2 = H^2 + B^2$

Where

H = Height of the building

L = Length of the first ladder = 20ft

B = Distance from the base of the building = 10ft

So, we have:

$20^2 = H^2 + 10^2$

$400 = H^2 + 100$

$H^2 = 400 - 100$

$H^2 = 300$

$H = \sqrt{300$

Next, is to determine the distance of the new ladder from the base of the building (B) using Pythagoras theorem using:

$L_2^2 = H^2 + B_2^2$

Where

$L_2 = 18$ --- Length of the second ladder

$H = \sqrt{300$ ----- Height of the building

So, we have:

$18^2 = \sqrt{300}^2 + B_2^2$

$324 = 300 + B_2^2$

$B^2_2 = 324 - 300$

$B^2_2 = 24$

Solving for B2, we have:

$B_2 = \sqrt{24$

$B_2 = 4.9$

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Angles 1 and 2 are called?

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3 years ago

Read 2 more answers

A sequence can be generated by using an=4a(n-1)

WITCHER [35]

This problem is half understanding the question and half just plugging in numbers.

Understanding the Question:
So its saying that n = whole number bigger than one, making n = 2, 3, 4, 5, etc.

Now try plugging any number n into the $a_n = 4a_{n-1}$ equation. Let's use 2 to make it simple. If you plug in 2, you get:
$a_2 = 4a_{2-1}\\
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Remember that the subscripts (the small numbers in the corner) are simply the term number. $a_1$ is term 1, $a_2$ is term 2, etc. The equation is saying that "term 2 = 4 times term 1."

If you keep plugging in random numbers for n, you'll see that the equation basically says, "a term = 4 times the term right before it." That makes it easy to plug and chug to get our answer.

Plugging and Chugging
Now we understand the question is asking for the first 4 terms, and each term is 4 times the last term. You're told that term 1, $a_1$ = 6. That makes the next four terms:

Term 2
$a_2 = 4a_{2-1}\\
 a_2 = 4a_1\\
 a_2 = 4(6)\\
 a_2 = 24$

Term 3
$a_3 = 4a_{3-1}\\ a_3 = 4a_2\\ a_3 = 4(24)\\ a_3 = 96$

Term 4
$a_4 = 4a_{4-1}\\ a_4 = 4a_3\\ a_4 = 4(96)\\ a_4 = 384$

Your first 4 terms are 4, 24, 96, and 384.

--------

Answer: A) F

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