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3 years ago

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3. What is the density of gold

Physics

1 answer:

Natali5045456 [20]3 years ago

5 0

Answer:

$19.3 g/cm^3$

Explanation:

The density of a material is given by

$\rho = \frac{m}{V}$

where

m is the mass of the sample

V is its volume

for the sample of gold in this problem,

m = 38.6 g

$V=2 cm^3$

Substituting into the formula, we find the density of gold:

$\rho = \frac{38.6}{2}=19.3 g/cm^3$

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What is the wavelength of an X-ray photon with energy 8.0 keV (8000 eV)? (1 eV = 1.60 × 10−19 joule.)

katrin2010 [14]

Answer:

Wavelength = 0.15 nm

Frequency = $1939.3939\times 10^{15}Hz$

Explanation:

We have given photon energy E = 8 KeV = 8000 eV

In question it is given that $1eV=1.6\times 10^{-19}J$

So $8000eV=1.6\times 8000\times 10^{-19}=12800\times 10^{-19}j$

Plank's constant $h=6.6\times 10^{-34}js$

We know that photon energy is given by $E=h\nu$

So $12800\times 10^{-19}=6.6\times 10^{-34}\nu$

$\nu =1939.3939\times 10^{15}Hz$

Now wavelength $\lambda =\frac{c}{f}=\frac{3\times 10^8}{1939.3939\times 10^{15}}=0.0015\times 10^{-7}m=0.15nm$

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4 years ago

In rural areas, water is often extracted from underground by pumps. Consider an underground water source whose free surface is 6

stealth61 [152]

Answer:

W = 9533.09 Watt

Explanation:

given,

diameter of pipe inlet, d₁ = 10 cm

r₁ = 5 cm

diameter of pipe outlet, d₂ = 15 cm

r₂= 7.5 cm

head upto water level is to rise = 60 + 5

= 65 m

flow rate = 0.015 m³/s

we know

A₁ v₁ = A₂ v₂ = Q

π r₁² v₁ = π r₂² v₂ = 0.015

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{7.5^2}{5^2} v_2$

$v_1= 2.25 v_2$

$v_2 = \dfrac{0.015}{\pi r_2^2}$

$v_2 = \dfrac{0.015}{\pi 0.075^2}$

v₂ = 0.848 m/s

v₁ = 1.908 m/s

Applying Bernoulli's equation

$P_p = \dfrac{1}{2}\rho (v_2^2-v_1^2)+ \rho g h$

$P_p= \dfrac{1}{2}\times 1000\times (0.848^2-1.908^2)+ 1000\times 9.8\times 65$

$P_p= 635539.32 Pa$

P_p is the pump pressure

Power of the pump

W = P_p x Q

W = 635539.32 x 0.015

W = 9533.09 Watt

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Answer:

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