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julia-pushkina [17]
4 years ago
6

Una barra metálica de 2 m de largo recibe una fuerza que lo provoca una alargamiento o variación en su longitud de 0.3 cm ¿Cuál

es el valor de la tensión unitaria o deformación lineal?
Physics
1 answer:
Elan Coil [88]4 years ago
3 0

Answer:

La deformación unitaria lineal experimentada por la barra es 1.5\times 10^{-3}.

Explanation:

De la Mecánica de Materiales sabemos que la deformación unitaria lineal es la razón de la variación de la longitud con respecto a su longitud inicial. Al asumirse que la variación longitudinal es muy pequeña con respecto a la longitud inicial, se puede utilizar la siguiente ecuación:

\epsilon = \frac{\Delta l}{l_{o}} (Eq. 1)

Donde:

\epsilon - Deformación unitaria, adimensional.

\Delta l - Cambio longitudinal, medido en metros.

l_{o} - Longitud inicial, medida en metros.

Si conocemos que \Delta l = 3\times 10^{-3}\,m y l_{o} = 2\,m, entonces la deformación unitaria lineal es:

\epsilon = \frac{3\times 10^{-3}\,m}{2\,m}

\epsilon = 1.5\times 10^{-3}

La deformación unitaria lineal experimentada por la barra es 1.5\times 10^{-3}.

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3 years ago
Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is t
den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

F\propto \dfrac{1}{r^2}

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N    

We need to find the electric force between charges if the new separation of 0.25 cm. So,

\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N

So, the new force is 64 N if the separation between charges is 64 N.

7 0
3 years ago
What is the volume of a 6-sided die a cube with each side measuring 1.6 cm across
Savatey [412]
4.096. You just have to multiply 1.6 three times. V=s^3
8 0
4 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

4 0
3 years ago
Many species cool themselves by sweating, because as the sweat evaporates, heat is transferred to the surroundings. A human exer
max2010maxim [7]

Answer:

70713

Explanation:

Because you need to multiply the amount of water lost (2430) by the time (29.1) which will equal 70713J/g needed to counter the loss.

Hope this helps:)

4 0
3 years ago
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