Answer: 7022.2kg/m³, yes, I was cheated
Explanation:
Density of an object is defined as the ratio of the mass of the object to its volume. Mathematically;
Density = Mass/Volume
Note that the unit of both mass and volume must be standard unit.
Given mass = 0.0158kg
Dimension of the metal = 5mm×15mm×30mm
Note that 1mm = 0.001m
The volume of the metal will be
0.005×0.015×0.03
= 0.00000225m³
Density = 0.0158/0.00000225
Average density of the metal = 7022.2kg/m³
Since the standard density of Gold is 19,320kg/m³ and is higher than the density prescribed for me, it shows the I was cheated.
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I believe this is right but double check to make sure :)
Answer:
increase
decrease
Explanation:
using formula
Vt=mg/6πηr
so if m increases V increases
r is the denominator so if r increases V decreases
The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is
<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J
That is,
• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point
• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium
so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.
By the work-energy theorem,
<em>W</em> = ∆<em>K</em> = <em>K</em>
where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So
<em>W</em> = 1/2 <em>mv</em> ²
where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get
<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s
Answer:
Minimum thickness; t = 9.75 x 10^(-8) m
Explanation:
We are given;
Wavelength of light;λ = 585 nm = 585 x 10^(-9)m
Refractive index of benzene;n = 1.5
Now, let's calculate the wavelength of the film;
Wavelength of film;λ_film = Wavelength of light/Refractive index of benzene
Thus; λ_film = 585 x 10^(-9)/1.5
λ_film = 39 x 10^(-8) m
Now, to find the thickness, we'll use the formula;
2t = ½m(λ_film)
Where;
t is the thickness of the film
m is an integer which we will take as 1
Thus;
2t = ½ x 1 x 39 x 10^(-8)
2t = 19.5 x 10^(-8)
Divide both sides by 2 to give;
t = 9.75 x 10^(-8) m
