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shtirl [24]
3 years ago
11

A chain 27 feet long whose weight is 95 pounds is hanging over the edge of a tall building and does not touch the ground. How mu

ch work is required to lift the entire chain to the top of the building?
Mathematics
1 answer:
bazaltina [42]3 years ago
6 0

Answer:

Work done = 2,565 lb-ft

Step-by-step explanation:

The density of the chain in lb/ft  is equal to

\frac{95}{27} lb/ft

Total Work done is equal to the summation of  work done up to the top of building.

We will use integration for this evaluation.

The mathematical relation is as follows

\int\limits^{40}_0 {X} \, pdx

where "p" is the density

Substituting the given values, we get

\frac{95}{27} \int\limits^{27}_0 {X} \, dx\\\frac{95}{27}  X^2\int\limits^{27}_0\\ \frac{95}{27}  {27^2 - 0^2}\\\frac{95}{27}  * 27 * 27\\95* 27\\= 2,565

Work done = 2,565 lb-ft

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The Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
kirza4 [7]

Answer:

Step-by-step explanation:

given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging

Clerk          I           II           III    total

   

Pack      0.33          0.23    0.44 1

   

Defective 0.02 0.025    0.015  

   

Pack&def 0.0066 0.00575 0.0066 0.01895

a)  probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate

= P(II clerk) -P(II clerk and defective) = 0.23-0.00575=0.22425

b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)

=0.01895

c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3

=P(clerk 3 and def)/P(defective)

=\frac{0.0066}{0.01895} \\=0.348285

8 0
3 years ago
If an object is launched at an angle of 28° with an initial velocity of 133 ft./s from a height of 6 feet how many seconds will
Tema [17]

Answer:

Hence after  3.98 sec  i.e  4 sec Object will hit the ground  .

Step-by-step explanation:

Given:

Height= 6 feet

Angle =28 degrees.

V=133 ft/sec

To Find:

Time in seconds after which it will hit the ground?

Solution:

<em>This problem is related to projectile motion for objec</em>t

First calculate the Range for object  and it is given by ,

R=v^2Sin(2Ф)/g

Here R= range  g= acceleration due to gravity =9.8 m/sec^2

1m =3.2 feet

So 9.8 m, equals to 9.8 *3.2=31.36 ft

So g=31.36 ft/sec^2. and 2Ф=2(28)=56

R=133^2*Sin(56)/31.36

R=14664.84/31.36

R=467.62  fts

Now using Formula for time and range as

R=VxT

Vx is horizontal velocity

Vx=V*cosФ

Vx=133*cos(28)

Vx=117.43  ft/sec

So above equation becomes as ,

467.62=117.43*T

T=467.62/117.43

T=3.98 sec

T is approximately equals to 4 sec.

4 0
3 years ago
Olivia planted 25 tomato plants, but only produced 20 tomatoes. What percentage of plants produce tomatoes?
solmaris [256]

Answer:

It would be 5.

Step-by-step explanation:

Hope this help!

8 0
3 years ago
When x=4 what is the value of 8x
Naddik [55]
8*4=32 :) I hope this helps!
5 0
3 years ago
Please help! Question posted above ^
emmainna [20.7K]

Answer:

( x^2 - 6 )^2

Step-by-step explanation:

  • x^4 -12x^2 +36
  1. Find one factor
  2. Factor by grouping
  • (x^2 - 6x) (x^2 - 6x)

*solution* ( x^2 - 6 )^2

8 0
3 years ago
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