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11Alexandr11 [23.1K]

3 years ago

14

In a lab, you perform a double slit experiment to measure the wavelength of a particular laser. When aimed at two slits spaced b

y 0.22 mm0.22mm an interference pattern is formed on a screen 0.5 m away. You measure that 10 bright lines of the interference pattern cover a total of 1.3 cm1.3cm width on the screen. What is the wavelength of the laser light?

1 answer:

bekas [8.4K]3 years ago

7 0

Answer:

The wavelength of the laser light is 572 nm.

Explanation:

Given that,

Distance between slit d= 0.22 mm

Distance of screen D= 0.5 m

Bright lines = 10

Width y= 1.3 cm

We need to calculate the wavelength

Using formula of young

$y=\dfrac{n\lambda D}{d}$

$\lambda=\dfrac{dy}{nD}$

Put the value into the formula

$\lambda=\dfrac{0.22\times10^{-3}\times1.3\times10^{-2}}{10\times0.5}$

$\lambda=5.72\times10^{-7}\ m$

$\lambda=572\times10^{-9}\ m$

Hence, The wavelength of the laser light is 572 nm.

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The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>

The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

$m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s$

The time taken for the puck to reach 15 m is calculated as follows;

$t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s$

The distance traveled by the hockey player at the calculated time is;

$d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m$

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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2 years ago

An airplane travels 4000m in 16 seconds on a heading of 35 whats is its velocity

Marizza181 [45]

In the question, you just gave a complete and detailed
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3 years ago

When two hydrogen nuclei combine to form one helium nucleus, nuclear fusion has taken place. Please select the best answer from

kirill115 [55]

The answer is true
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3 years ago

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The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

Nadusha1986 [10]

Answer:

The charge resides on the outer surface = $1.245 \times 10^{-12}$ C

Explanation:

Surface area of cell $(A) = 4.3\times 10^{-9} m^{2}$

Separation between two plate $(d) = 1.1 \times 10^{-8} m$

Dielectric constant $(k) = 4.2$

Potential difference $(\Delta V) = 85.7 \times 10^{-3} V$

The capacitance of parallel plate capacitor in free space is given by,

$C = \frac{\epsilon_{o} A }{d}$

Where $\epsilon_{o} =$ permittivity of free space = $8.85 \times 10^{-12}$

The Capacitance of capacitor is increase by $k$ times when it placed in dielectric medium.

$C_{dielectric} = \frac{k \epsilon_{o} A }{d}$

And we know that, $C = \frac{Q}{ \Delta V}$

So charge on the outer surface is given by,

$Q = \frac{k \epsilon A \Delta V }{d}$

$Q = \frac{4.2 \times 4.3 \times 10^{-9} \times 8.85 \times 10^{-12} \times 85.7\times 10^{-3} }{1.1 \times 10^{-8} }$

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3 0

4 years ago

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

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4 years ago