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Verdich [7]
3 years ago
13

The atomic radius of metal X is 1.20 × 102 picometers (pm) and a crystal of metal X has a unit cell that is face-centered cubic.

Calculate the density of metal X (atomic weight = 42.3 g/mol).
Physics
1 answer:
dmitriy555 [2]3 years ago
4 0
The density of the metal can be determined through the formula [n*MW]/ Na*[a^3] . substituting, we get, 

<span>d = [n*MW]/ Na*[a^3] 
</span><span>d = [4 atoms*42.3 g/mol]/ [6.022 x 1023atoms/mol* (sqrt 8 *1.20x10-10)^3] 
</span>d = 0.719 g/cm3
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Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Bas_tet [7]

Given info

d = 0.000250 meters = distance between slits

L = 302 cm = 0.302 meters = distance from slits to screen

\theta_8 = 1.12^{\circ} = angle to 8th max (note how m = 8 since we're comparing this to the form \theta_m)

x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters} (n = 5 as we're dealing with the 5th minimum )

---------------

Method 1

d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}

Make sure your calculator is in degree mode.

-----------------

Method 2

\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\

-----------------

Method 3

\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\

There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.

7 0
3 years ago
A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so
joja [24]

ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em>          l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound  intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em>     </em>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

3 0
3 years ago
A 10-kilogram bicycle is traveling at a speed of 2m/s. what is the bike's kinetice energy?
madam [21]
The kinetic energy of an object of mass m moving with speed v is given by:
K= \frac{1}{2}mv^2
For the bicycle in our problem, m=10 kg and v=2 m/s, so the kinetic energy is
K= \frac{1}{2}(10 kg)(2 m/s)^2=20 J
4 0
3 years ago
Yes this nees helppppppppppppppp
kozerog [31]

Answer:

lol

Explanation:

it was funny

6 0
3 years ago
Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
USPshnik [31]

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

<h3>Time of motion of the hammer</h3>

The time of motion is calculated as follows;

t = √(2h/g)

where;

  • h is height of fall
  • g is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

8 0
2 years ago
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