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vodomira [7]
3 years ago
5

Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a

uniform glass rod?​
Physics
1 answer:
Mice21 [21]3 years ago
5 0

Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

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If the capacity of the lungs of an adult were 3.8 L, how many moles of air would the lungs contain at body temperature and atmos
Ahat [919]

Answer:

  • 0.14955  mol

Explanation:

We can solve this problem using the ideal gas law

P \ V = \ n \ R \ T

where P is the pressure, V the volume, n the number of moles, R the ideal gas constant and T the temperature.

We can use the atmospheric pressure as 1 atm, and the body temperature as 36.5 °C, in Kelvin this is:

T_{body} = 36.5 \° C = (36.5 + 273.15) K = 309.65 \ K

The ideal gas constant is:

R = 0.082057 \frac{L \ atm}{ K \ mol}

taking all this in consideration, the number of moles will be:

n = \frac{P \ V}{  R \ T }

n = \frac{1 \ atm * 3.8 \ L  }{ 0.082057 \frac{L \ atm}{ K \ mol} *  309.65 \ K } [/tex]

n = 0.14955 \ mol

3 0
3 years ago
A student calculates the density of a copper cube to be 4.15 g/cm . If the accepted value is 8.64 g/cm the percentage error in h
Helen [10]

 The percentage error in his experimental value is -51.97%.

<h3>What is percentage error?</h3>

This is the ratio of the error to the actual measurement, expressed in percentage.

To calculate the percentage error of the student, we use the formula below.

Formula:

  • Error(%) = (calculated value-accepted value)100/(accepted............. Equation 1

From the question,

Given:

  • Calculated value = 4.15 g/cm
  • accepted value = 8.64 g/cm

Substitute these values into equation 1

  • Error(%) = (4.15-8.64)100/8.64
  • Error(%) = -4.49(100)/8.64
  • Error(%) = -449/8.64
  • Error(%) = -51.97 %

 

Hence, The percentage error in his experimental value is -51.97%.

Learn more percentage error here: brainly.com/question/5493941

8 0
3 years ago
Even through there is equal and opposite reaction,usually the two forces are not seen balanced.Why?
Tomtit [17]

Answer:

This may refer to a situation like:

"one person pushes a box, if there is equal and opposite reaction why the box moves and the person does not?"

Remember the second Newton's law:

F = m*a

suppose that the mass of the person is 3 times the mass of the box.

So, if the box has a mass M, the person will have a mass 3*M

Then the Newton's equation for the box when the person pushes with a force F is:

F = M*a

solving for the acceleration, we get:

F/M = a

While the person is also pushed by the box with a force with the same magnitude, then the equation for the person is:

F = (3*M)*a'

Solving for the acceleration, we get:

F/(3M) = a'

Now we can compare the acceleration of the box (F/M) with the acceleration of the person (F/3M).

Is easy to see that the acceleration of the box is 3 times the acceleration of the person.

So regardless of the fact that both the box and the person experience a force with the same magnitude, the box will move more due to this force.

This is why in situations like this, the forces do not seem balanced.

5 0
3 years ago
Convert 2536 mm/min to m/s. Use dimensional analysis.
Deffense [45]

Answer:

<u>As</u><u> </u><u>we</u><u> </u><u>kno</u><u>w</u><u> </u><u>that</u><u>,</u><u> </u>

  • 1 mm/min = 1.66667E-5 m/s
  • 1 m/s = 60000 mm/min

<u>Now</u><u>,</u><u> </u><u>come</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>question</u><u> </u><u>-</u><u> </u>

\\  \implies \sf \: 2536 \times  1.66667E-5 m/s \\  \\  \\  \implies \sf \blue{0.0422666667 \:m/s  } \\

Result : 2536 mm/min = 0.0422666667 m/s.

6 0
4 years ago
The value of g is greater at the poles than at the equator why ​
babunello [35]

Answer:

because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

6 0
3 years ago
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