Which can occur during a physical change? (SELECT ALL THAT APPLY!.)

Older railroad tracks in the U.S. are made of 12-m-long pieces of steel. When the tracks are laid, gaps are left between section

____ [38]

Answer: ∆L = 0.49cm ≈ 0.50cm

Therefore there should be 0.5 cm gap between each piece of steel.

Explanation:

Thermal expansion of steel is the increase in size of steel as a result of increased temperature. It can be represented by the mathematical expression:

∆L = L(k)∆T .....1

Where;

∆L is the change in length

L is the initial length

∆T is the change in temperature

k is the specific Linear expansion coefficient.

Given;

L = 12m

∆T = 50°C - 16°C = 34°C

k (for steel) = 1.2 × 10^-6 /C

Substituting the values into the equation 1

∆L = 12 × 34 × 12×10^-6

∆L = 4896 × 10^-6 m

∆L = 0.49cm ≈ 0.50cm

Therefore there should be 0.5 cm gap between each piece of steel.

6 0

3 years ago

Read 2 more answers

b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

Using the result of the preceding problem, (a) calculate the distance between fringes for 633-nm light falling on double slits s

vekshin1

To solve this problem it is necessary to apply the concepts related to the concept of superposition and the fringe separation for double slit experiment.

The equation can be written as

$\Delta y = \frac{x\lambda}{d}$

Where

$\Delta y$ = Distance between fringes

x = distance between slits and screen

d = Distance between slits

$\lambda$ = Wavelength

Our values are given as

d= 0.08mm

x =3m

$\lambda = 633nm$

In this way replacing in the equation,

$\Delta y = \frac{x\lambda}{d}$

$\Delta y = \frac{3m(633nm*(\frac{1*10^{-9}}{1nm}))}{0.08*(\frac{1*10^{-3}m}{1mm})}$

$\Delta y = 2.37*10^{-2}m$

$\Delta y = 2.37cm$

Therefore the distance between the fringes is 2.37cm

PART B) For the case in which it is submerged in water it is necessary to apply the relationship of the fringes with the index of refraction therefore

$\Delta Y_2 = \frac{\Delta Y_1}{n}$