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Stolb23 [73]
3 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]3 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>
TRuthS2 years ago
0 0

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

TRuthS
2 years ago
then, minus 425 x 1.8 x 66 = 224 680.5 J
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2 years ago
1. A farmer divided a field into 1-foot by 1-foot sections and tested soil samples from 32 randomly selected sections in the fie
Ad libitum [116K]
Part A

Answers:
Mean = 5.7
Standard Deviation = 0.046

-----------------------

The mean is given to us, which was 5.7, so there's no need to do any work there.

To get the standard deviation of the sample distribution, we divide the given standard deviation s = 0.26 by the square root of the sample size n = 32
So, we get s/sqrt(n) = 0.26/sqrt(32) = 0.0459619 which rounds to 0.046

================================================

Part B

The 95% confidence interval is roughly (3.73, 7.67)
The margin of error expression is z*s/sqrt(n)
The interpretation is that if we generated 100 confidence intervals, then roughly 95% of them will have the mean between 3.73 and 7.67

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*5.7/sqrt(32)
ME = 1.974949
The margin of error is roughly 1.974949

The lower and upper boundaries (L and U respectively) are:
L = xbar-ME
L = 5.7-1.974949
L = 3.725051
L = 3.73
and
U = xbar+ME
U = 5.7+1.974949
U = 7.674949
U = 7.67

================================================

Part C

Confidence interval is (5.99, 6.21)
Margin of Error expression is z*s/sqrt(n)
If we generate 100 intervals, then roughly 95 of them will have the mean between 5.99 and 6.21. We are 95% confident that the mean is between those values.

-----------------------

At 95% confidence, the critical value is z = 1.96 approximately

ME = margin of error
ME = z*s/sqrt(n)
ME = 1.96*0.34/sqrt(34)
ME = 0.114286657
The margin of error is roughly 0.114286657

L = lower limit
L = xbar-ME
L = 6.1-0.114286657
L = 5.985713343
L = 5.99

U = upper limit
U = xbar+ME
U = 6.1+0.114286657
U = 6.214286657
U = 6.21
5 0
3 years ago
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