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pav-90 [236]
3 years ago
15

Dos láminas, una de cobre y otra de hierro, se encuentran soldadas y empotradas enuna pared como lo muestra la figura 13. Si las

láminas se encuentra a 20 ºC ysabiendo que el coeficiente de dilatación térmica del cobre es mayor que el hierro,entonces se podría predecir que a una temperatura de 100 ºC
A) El extremo libre se doblará hacia A.
B) el extremo libre se doblará hacia B.
C) las láminas se dilatarán sin doblarse.
D) las láminas se contraerán sin doblarse.
E)la lámina de coeficiente de dilatación térmica menorimpedirá la dilatación de la otra.
Physics
1 answer:
PolarNik [594]3 years ago
7 0

Answer:

umm i gotta put this in English!!!!!

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6 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

          $= 2 \times \sqrt{\frac{100}{4}}$

          = 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

    m' = 2m

    Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

  $=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0
3 years ago
10 turns of wire are closely wound around a pencil as shown in the figure. when measured using a scale as shown, the length of t
Mila [183]

Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = \frac{length of total turns}{number of turns}

                           = \frac{25}{10}

                           = 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x  = \frac{2.5}{10}

   = 0.25 cm

The wire is 0.25 cm thick.

8 0
2 years ago
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