PLEASE HELP HURRY!! What is the value of x?

A uniform beam of length L = 7.30m and weight = 4.45x10²N is carried by two ovorkers , Sam and Joe - Determine the force exert e

Mama L [17]

Answer:

Effort and distance = Load x distance

7.30 x 4.45x10^2N = 3.2485 X 10^3N

We then know we can move 3 points to the right and show in regular notion.

= 3248.5

Divide by 2 = 3248.5/2 = 1624.25 force

Step-by-step explanation:

In the case of a Second Class Lever as attached diagram shows proof to formula below.

Load x distance d1 = Effort x distance (d1 + d2)

The the load in the wheelbarrow shown is trying to push the wheelbarrow down in an anti-clockwise direction whilst the effort is being used to keep it up by pulling in a clockwise direction.

If the wheelbarrow is held steady (i.e. in Equilibrium) then the moment of the effort must be equal to the moment of the load :

Effort x its distance from wheel centre = Load x its distance from the wheel centre.

This general rule is expressed as clockwise moments = anti-clockwise moments (or CM = ACM)

This gives a way of calculating how much force a bridge support (or Reaction) has to provide if the bridge is to stay up - very useful since bridges are usually too big to just try it and see!

The moment of the load on the beam (F) must be balanced by the moment of the Reaction at the support (R2) :

Therefore F x d = R2 x D

It can be seen that this is so if we imagine taking away the Reaction R2.

The missing support must be supplying an anti-clockwise moment of a force for the beam to stay up.

The idea of clockwise moments being balanced by anti-clockwise moments is easily illustrated using a see-saw as an example attached.

We know from our experience that a lighter person will have to sit closer to the end of the see-saw to balance a heavier person - or two people.

So if CM = ACM then F x d = R2 x D

from our kitchen scales example above 2kg x 0.5m = R2 x 1m

so R2 = 1m divided by 2kg x 0.5m

therefore R2 = 1kg - which is what the scales told us (note the units 'm' cancel out to leave 'kg')

But we can't put a real bridge on kitchen scales and sometimes the loading is a bit more complicated.

Being able to calculate the forces acting on a beam by using moments helps us work out reactions at supports when beams (or bridges) have several loads acting upon them.

In this example imagine a beam 12m long with a 60kg load 6m from one end and a 40kg load 9m away from the same end n- i.e. F1=60kg, F2=40kg, d1=6m and d2=9m

CM = ACM

(F1 x d1) + (F2 x d2) = R2 x Length of beam

(60kg x 6m) + (40Kg x 9m) = R2 x 12m

(60kg x 6m) + (40Kg x 9m) / 12m = R2

360kgm + 360km / 12m = R2

720kgm / 12m = R2

60kg = R2 (note the unit 'm' for metres is cancelled out)

So if R2 = 60kg and the total load is 100kg (60kg + 40kg) then R1 = 40kg

4 0

3 years ago

Find the exact value of y. V6 45°

Lynna [10]

Answer:

more info

Step-by-step explanation:

4 0

3 years ago

Nineteen immigrants to the U.S. were asked how many years, to the nearest year, they have lived in the U.S. The data are as foll

JulsSmile [24]

Answer:

a) The frequency of the data "15" and "20" is 2 for both, not 1; this means their relative frequency is 2/19 for both, not 1/19; finally, the cumulative relative frequency in the row of the data "15" should be 0.8947, not 0.8421. This error might have happened because someone didn't count the numbers correctly, so they only noticed one "15" and one "20" when, in fact, there were two people that had lived in the U.S. for 15 years, and two more people for 20 years. On the other hand, the error in the cumulative relative frequency happened because it accounted for only one person living in the U.S. for 15 years, instead of two people.

b) Roughly 47% of the people surveyed have lived in the U.S. from 0 to 5 years, not for 5 years. The cumulative relative frequency in this row (47%) accounts for every data gathered so far, not just the "5 years" row. The correct statement would be that 3 out of 19, or 15.8% (relative frequency) of the people surveyed have lived in the U.S. for 5 years.

Step-by-step explanation:

1) First of all, to avoid errors like the one in the problem's table, we should first place the given numbers from least to greatest, so we can construct a new frequency table by ourselves. Let's do just that, and we'll end up with something like this:

0 , 0, 2 , 2, 2, 4, 5, 5, 5, 7, 7, 10, 10, 12, 12 , 15, 15, 20, 20

Now we'll have a much easier time from now on.

2) The second step is to construct the Data and Frequency columns. Just place each unique integer in a new row of the Data column, then count how many times that unique integer was found, and, finally, place that number below the Frequency column (Please refer to the Excel Worksheet provided as an attachment). 

Let's do it as follows:

Data Frequency

0 2

2 3

4 1

5 3

7 2

10 2

12 2

15 2

20 2

Note that we counted "15" and "20" twice! So each one of those rows have a frequency of 2, not 1 as the table presented in the problem suggests.

3) Next, we want to construct the Relative frequency and Cumulative relative frequency columns. For the relative frequency column, we just divide the frequency of each row by the total number of immigrants surveyed, which is 19. For the cumulative relative frequency column, we will get each row's relative frequency, and add the cumulative relative frequency of the row before it. Note that for the first row, the cumulative relative frequency is the same as its relative frequency.

We should get something like this:

Data Frequency Relative frequency Cumulative relative frequency

0 2 2/19 0.1053

2 3 3/19 0.2632

4 1 1/19 0.3158

5 3 3/19 0.4737

7 2 2/19 0.5789

10 2 2/19 0.6842

12 2 2/19 0.7895

15 2 2/19 0.8947

20 2 2/19 1.0000

Note that the relative frequency for both "15" and "20" is 2/19 instead of 1/19! Also, we got a cumulative relative frequency of 0.8947 in the row of "15", instead of 0.8421.

4) a) We have just fixed the error in the table, but we have to explain how someone might have arrived at the incorrect number(s). The most logical way that someone might have gotten the incorrect frequencies of "15" and "20" is that they didn't count the numbers correctly while building the Frequency column. This could have happened because that person probably didn't order the numbers from least to greatest, as we did in Step 1, which makes it way easier to get the frequency of each data without making a mistake.

5) b) We have now to explain what is wrong with the statement "47% of the people surveyed have lived in the U.S. for 5 years.

To answer that, we can refer to the relative frequency of the row of the data "5", which tells us that 3 out of 19 (or roughly 15.8%) of the people surveyed have lived in the U.S. for 5 years. Relative frequency is telling us the percentage of people that have lived for this amount of time.

By contrast, the cumulative relative frequency of this same row tells us that 0.4737, or roughly 47%, of the people surveyed have lived for 5 years or less. Cumulative relative frequency accounts for the data presented in its row, plus the data presented in the rows before it.

So the correct statement would be either: