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 (
), en metros:
![L = L_{o}\cdot [1+\alpha\cdot (T_{f}-T_{o})]](https://tex.z-dn.net/?f=L%20%3D%20L_%7Bo%7D%5Ccdot%20%5B1%2B%5Calpha%5Ccdot%20%28T_%7Bf%7D-T_%7Bo%7D%29%5D)
(1)
Donde:
- Longitud inicial del puente, en metros. 
- Coeficiente de dilatación, sin unidad.
- Temperatura inicial, en grados Celsius.
- Temperatura final, en grados Celsius.
Si tenemos que 
, 
, 
y 
, 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)]](https://tex.z-dn.net/?f=L%20%3D%20%28100%5C%2Cm%29%5Ccdot%20%5B1%2B%2811.5%5Ctimes%2010%5E%7B-6%7D%29%5Ccdot%20%2824%5C%2C%5E%7B%5Ccirc%7DC%20-%208%5C%2C%5E%7B%5Ccirc%7DC%29%5D)
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
Answer:
(a) 9.36 kHz
(b) 3.12 kHz
Explanation:
(a)
V = speed of sound
= speed of airplane = (0.5) V
f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz
f' = Frequency heard by the stationary listener
Using Doppler's effect
f' = 9360 Hz
f' = 9.36 kHz
(b)
V = speed of sound
= speed of airplane = (0.5) V
f = actual frequency of sound emitted by airplane = 4.68 kHz = 4680 Hz
f' = Frequency heard by the stationary listener
Using Doppler's effect
f' = 3120 Hz
f' = 3.12 kHz
Answer:
Approximately 
assuming that the effect of gravity on the box can be ignored.
Explanation:
If the force 
is constant, then the work would be found with 
. However, this equation won't work for this question since the
,
For this particular question, 
and 
. Apply this equation:
![\begin{aligned}W &= \int\limits_{x_0}^{x_1} F\, d x \\ &= \int\limits_{0\; \rm m}^{14.0\; \rm m} \left[{18.0\; \rm N} - {\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x \right]\, d x \\ &= \left[{(18.0\; \rm N)}\cdot x - \frac{1}{2}\;{\left(0.530\; {\rm N \cdot m^{-1}}\right)}\cdot x^2\right]_{x = 0\; \rm m}^{x = 14.0\; \rm m} \approx 200.06\; \rm N \cdot m\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7DW%20%26%3D%20%5Cint%5Climits_%7Bx_0%7D%5E%7Bx_1%7D%20F%5C%2C%20d%20x%20%5C%5C%20%26%3D%20%5Cint%5Climits_%7B0%5C%3B%20%5Crm%20m%7D%5E%7B14.0%5C%3B%20%5Crm%20m%7D%20%5Cleft%5B%7B18.0%5C%3B%20%5Crm%20N%7D%20-%20%7B%5Cleft%280.530%5C%3B%20%7B%5Crm%20N%20%5Ccdot%20m%5E%7B-1%7D%7D%5Cright%29%7D%5Ccdot%20x%20%20%5Cright%5D%5C%2C%20d%20x%20%5C%5C%20%26%3D%20%5Cleft%5B%7B%2818.0%5C%3B%20%5Crm%20N%29%7D%5Ccdot%20x%20-%20%5Cfrac%7B1%7D%7B2%7D%5C%3B%7B%5Cleft%280.530%5C%3B%20%7B%5Crm%20N%20%5Ccdot%20m%5E%7B-1%7D%7D%5Cright%29%7D%5Ccdot%20x%5E2%5Cright%5D_%7Bx%20%3D%200%5C%3B%20%5Crm%20m%7D%5E%7Bx%20%3D%2014.0%5C%3B%20%5Crm%20m%7D%20%5Capprox%20200.06%5C%3B%20%5Crm%20N%20%5Ccdot%20m%5Cend%7Baligned%7D)
.
(Side note: keep in mind that 
.)
Since friction is ignored, all these work should have been converted to the mechanical energy of this object.
Assume that the effect of gravity on this box can also be ignored. That way, there won't be a change in the gravitational potential energy of this object. Hence, all these extra mechanical energy would be in the form of the kinetic energy of this box.
That is:
.
Keep in mind that the kinetic energy of an object of mass 
and speed is:
.
Therefore:
.
Answer:
NICE HANDWRITING. I've done this before and I'm sure everything is correct. Although I would re-check to see if you had any mistakes with your calculations. Good job!
Explanation:
Answer:
E=mc2
Explanation:
It explains that in their interiors, atoms fuse together, creating energy.
