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gregori [183]
3 years ago
11

Does a calculation lose any information when converted from centimeters to meters? Explain

Physics
1 answer:
german3 years ago
7 0

Answer:

No

Explanation:

There are many units that are used to measure length of an object. For example centimeters, meters, millimeters etc.

There is a relationship between any of two units to measure lengths. If we want to convert some length from cm to m, it can be done as follows :

1 cm = 0.01 m

or

1 m = 100 cm

When we use this conversion, the calculation remains the same. Only the way to represent it will be different.

Hence, there is no lose of information when converted from centimeters to meters.

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After a displacement of 17 m, a train on a straight track is at the position xf = –2.5 m
EastWind [94]

-19.5m

-19.5+17=-2.5m

5 0
3 years ago
Read 2 more answers
What happens to the gravitational potential energy between two particles if the distance between them is halved? (a) It does not
mr_godi [17]

Answer:

The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).

Explanation:

The gravitational force is the force of mutual attraction that two objects with mass experience.

The Law of Universal Gravitation enunciated by Newton says that every material particle attracts any other material particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance that separates them. Mathematically this is expressed as:

F=G*\frac{m1*m2}{r^{2} }

where m1 and m2 are the masses of the objects, r the distance between them and G a universal constant that receives  the name of constant of gravitation.

If the distance between two particles is reduced by half, then, where F' is the new value of the gravitational force:

F'=G*\frac{m1*m2}{(\frac{r}{2} )^{2} }

F'=G*\frac{m1*m2}{\frac{(r )^{2} }{2^{2} } }

F'=G*\frac{m1*m2}{\frac{(r )^{2} }{4} }

F'=4*G*\frac{m1*m2}{r^{2} }

F'=4*F

<u><em> The gravitational potential energy between two particles, if the distance between them is halved, is multiplied by 4 (option c).</em></u>

7 0
3 years ago
Which is one way scientists indicate how precise and accurate there experimental measurements are
kherson [118]
They do the method 3 times to be sure. Because if you do it once, that could mean anything. If you do it twice, it may or may not have the same result. If you do it 3 times and it matches one of the previous answers, then it's likely that it's correct.
8 0
3 years ago
Which of the following is not a reason fluorescent lamps are advantages over incandescent lamps?
iren2701 [21]

It’s because flourecent lights operate at higher temperatures than incadecent lights.

3 0
3 years ago
A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in
lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

        A = 1.5 \times 2.30

           = 3.45 in^{2}

The given data is as follows.

          Allowable stress = 14,500 psi

          Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

           P_{max} = \sigma_{a}A

                       = 14500 \times 3.45

                       = 50025 lb

Now, maximum load from shear stress is as follows.

           P_{max} = 2 \times \tau_{a} \times A

                      = 2 \times 7100 \times 3.45

                      = 48990 lb

Hence, P_{max} will be calculated as follows.

       P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})

                  = 48990 lb

Thus, we can conclude that the maximum permissible load P_{max} is 48990 lb.

4 0
4 years ago
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