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tiny-mole [99]
3 years ago
15

In what way are liquids different from solids?.

Physics
1 answer:
puteri [66]3 years ago
7 0

Answer:

solid molecules are packed closely and liquid molecules are losely packed

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Remote Scottish island
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isotopes are the same element, but have different numbers of neutrons (but still have the same number of electrons and protons), hence have a different mass number.

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A student walks 31 m east, then 16 m west. Make east positive. What is their<br> displacement?
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Displacement = 31 - 16 = +15 m
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Un puente de acero de 100 m de largo a 8° C aumenta su temperatura a 24°C ¿Cuánto medirá su longitud? Valor del coeficiente de d
BartSMP [9]

La longitud <em>final</em> del puente de acero es 100.018 metros.

Asumamos que la dilatación <em>térmica</em> experimentada por el puente de acero es <em>pequeña</em>, de modo que podemos emplear la siguiente aproximación <em>lineal</em> para determinar la longitud <em>final</em> del puente de acero (L), en metros:

L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})] (1)

Donde:

  • L_{o} - Longitud inicial del puente, en metros.
  • \alpha - Coeficiente de dilatación, sin unidad.
  • T_{o} - Temperatura inicial, en grados Celsius.
  • T_{f} - Temperatura final, en grados Celsius.

Si tenemos que L_{o} = 100\,m, \alpha = 11.5\times 10^{-6}, T_{o} = 8\,^{\circ}C y T_{f} = 24\,^{\circ}C, entonces la longitud final del puente de acero es:

L = (100\,m)\cdot [1+(11.5\times 10^{-6})\cdot (24\,^{\circ}C - 8\,^{\circ}C)]

L = 100.018\,m

La longitud <em>final</em> del puente de acero es 100.018 metros.

Para aprender más sobre dilatación térmica, invitamos cordialmente a ver esta pregunta verificada: brainly.com/question/24953416

5 0
2 years ago
The cross section of a copper strip is 1.2 mmthick and 20 mm wide. There is a 25-A current through this cross section, with the
Naily [24]

To solve this problem it is necessary to use the concepts related to the Hall Effect and Drift velocity, that is, at the speed that an electron reaches due to a magnetic field.

The drift velocity is given by the equation:

V_d = \frac{I}{nAq}

Where

I = current

n = Number of free electrons

A = Cross-Section Area

q = charge of proton

Our values are given by,

I = 25 A

A= 1.2*20 *10^{-6} m^2

q= 1.6*10^{-19}C

N = 8.47*10^{19} mm^{-3}

V_d =\frac{25}{(1.2*20 *10^{-6})(1.6*10^{-19})(8.47*10^{19} )}

V_d = 7.68*10^{-5}m/s

The hall voltage is given by

V=\frac{IB}{ned}

Where

B= Magnetic field

n = number of free electrons

d = distance

e = charge of electron

Then using the formula and replacing,

V=\frac{(2.5)(25)}{(8.47*10^{28})(1.6*10^{-19})(1.2*10^{-3})}

V = 3.84*10^{-6}V

5 0
3 years ago
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