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Stolb23 [73]
2 years ago
10

A crane lifts a 425 kg steel beam vertically a distance of 66 m. How much work does the crane do on the beam if the beam acceler

ates upward at 1.8 m/s2? Neglect frictional forces ...?
Mathematics
2 answers:
Fiesta28 [93]2 years ago
6 0
<span>The solution to the problem is as follows:

W=Fd=mad=425*1.8*66 =50490 J

Therefore, it takes </span><span>50490 J of work the crane will do to accelerate the beam upward at 1.8 m/s^2.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

</span>
Guest1 year ago
0 0

425 x 66 x 9.81 = 275170.5 J = 2.75 X 10 ^5 J = 275.17 kJ

Guest
1 year ago
then, minus 425 x 1.8 x 66 = 224 680.5 J
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Hannah's school hosted a book donation. There are 150 students at her school, and they donated a total of b books! Hannah donate
svetoff [14.1K]

Answer:

x=3*\dfrac{b}{150}

Step-by-step explanation:

Let x = number of books that Hannah donated

We know that the number of students is 150 and the total number of books is represented by b, we can write that as:

Average number of books donated by each student = \dfrac{b}{150}

Now we know that Hannah donated 3 times as much as the average. We can write that as:

x=3*\dfrac{b}{150}

4 0
2 years ago
There are 14 dogs in a particular class at a dog show. Blue, red, yellow, and white ribbons will be awarded to the first through
stira [4]
There are 14 possibilities for first place.
There are 13 possibilities for second place, since a dog can't get both first and second place.
There are 12 possibilities for third place, since a dog can't get both first, second, and third place.
There are 11 possibilities for fourth place, since a dog can't get both first, second, third, and fourth place.

Multiply them together and get 24024 possibilities.

Hope this helps!

4 0
2 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
2 years ago
Solve the following addition and subtraction problems. a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam b. 5 sq.km 95 ha 8,994 sq.m + 11 sq.
kipiarov [429]
As a general rule to solve the problem we are going to transform all values to the lower unit.  
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  b. 5 sq.km 95 ha 8,994 sq.m + 11 sq. km. 11 ha 9,010 sq. m.
  5,000,000 sq m 95,0000 sq m 8,994 sq m + 11,000,000 sq m 110,000 sq
9,010 sq m
  5,103,994 sq m + 11,119,010 sq m = 16,223,004 sq m 
 c. 44 m – 5 dm
  44 m - 0.5 m = 43.5 m 
 d. 73 km 47 hm 2 dam - 11 km 55 hm

  73,000 m 4,700 m 20 m - 11,000 m 5,500 m 
77,720 m - 16,500 m = 61,220 m
5 0
2 years ago
Last Saturday, there were 1486 people at the Cineplex. There were about the same number of people in each of the 6 theaters. Bet
grigory [225]
<span>247 and 248 I.e. 247 in two thetheatres and 248 in the other 4. Hence 1486</span>
5 0
1 year ago
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