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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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How is a front defined in terms of weather ?. A) the leading edge of an anticyclone flow. B) the trailing edge of a warm front.
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6 0
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Suppose you have two small pith balls that are 5.5 cm apart and have equal charges of -29 nc?
Alina [70]
The question is missing, however, I guess the problem is asking for the value of the force acting between the two balls.

The Coulomb force between the two balls is:
F= k_e \frac{ q_1 q_2}{r^2}
where k_e=8.99\cdot10^9~N m^2 C^{-2} is the Coulomb's constant, q_1=q_2=29~nC=29\cdot 10^{-9}~C is the intensity of the two charges, and r=5.5~cm=0.055~m is the distance between them.

Substituting these numbers into the equation, we get
F=2.5~10^{-3}~N

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6 0
3 years ago
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sleet_krkn [62]
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3 0
3 years ago
A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.
Anna007 [38]

Answer:

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

N - mg = ma

so normal force on the block is given as

N = mg + ma

now the magnitude of kinetic friction on the block is given as

F_f = \mu N

F_f = \mu (mg + ma)

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

a = - \frac{F_f}{m}

a = -\mu_k(g + a)

now we know that bag hits the opposite wall at L distance away in time t

so we have

d = v t + \frac{1}{2}at^2

L = vt - \frac{1}{2}(\mu_k)(g + a) t^2

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

8 0
3 years ago
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