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3 years ago

How do I remember the Mohs hardness scale?

1 answer:

6 0

It's a hardness scale from 1-10 determining how easy or hard it is to scratch the mineral.

Remember that talc (like chalk or baby powder) is the softest and easiest to scratch then diamond being 10 is the hardest mineral to scratch or break or cut

You might be interested in

Calculate the wavelength of a photon having 3.26 x 10^-19 joules of energy

bekas [8.4K]

The energy of a photon is given by:
$E=hf$
where h is the Planck constant and f is the photon frequency.
We know the energy of the photon, $E=3.26 \cdot 10^{-19} J$ , so we can rearrange the equation to calculate the frequency of the photon:
$f= \frac{E}{h}= \frac{3.26 \cdot 10^{-19}J}{6.6 \cdot 10^{-34}Js}=4.94 \cdot 10^{14}Hz$

And now we can use the following relationship between frequency f, wavelength $\lambda$ and speed of light c to find the wavelength of the photon:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{4.94 \cdot 10^{14} Hz}=6.07\cdot 10^{-7} m=607 nm$

8 0

3 years ago

The average period of pendulum clock is found to be 1.2s at sea level. The period of the same pendulum on a mountain top is foun

Kipish [7]

Answer:

g' = 10.12m/s^2

Explanation:

In order to calculate the acceleration due to gravity at the top of the mountain, you first calculate the length of the pendulum, by using the information about the period at the sea level.

You use the following formula:

$T=2\pi \sqrt{\frac{l}{g}}$ (1)

l: length of the pendulum = ?

g: acceleration due to gravity at sea level = 9.79m/s^2

T: period of the pendulum at sea level = 1.2s

You solve for l in the equation (1):

$l=\frac{gT^2}{4\pi^2}\\\\l=\frac{(9.79m/s^2)(1.2s)^2}{4\pi^2}=0.35m$

Next, you use the information about the length of the pendulum and the period at the top of the mountain, to calculate the acceleration due to gravity in such a place:

$T'=2\pi \sqrt{\frac{l}{g'}}\\\\g'=\frac{4\pi^2l}{T'^2}$

g': acceleration due to gravity at the top of the mountain

T': new period of the pendulum

$g'=\frac{4\pi^2(0.35m)}{(1.18s)^2}=10.12\frac{m}{s^2}$

The acceleration due to gravity at the top of the mountain is 10.12m/s^2

5 0

3 years ago

What is the kinetic energy of a 8kg cat running 5m/s?

ivanzaharov [21]

<span>K.E = 0.5 * m * v^2 ( m = mass(Kg), V = Velocity(m/s)
= 0.5 * 8 * 5^2
= 4 * 25
= 100 J </span>

6 0

4 years ago

Astronauts use a centrifuge to simulate the acceleration of a rocket launch. The centrifuge takes 40.0 s to speed up from rest t

Vinvika [58]

Answer

Time period T = 1.50 s

time t = 40 s

r = 6.2 m

Angular speed ω = 2π/T

= $\dfrac{2\pi }{1.5}$

= 4.189 rad/s

Angular acceleration α = $\dfrac{\omega}{t}$

= $\dfrac{4.189}{40}$

= 0.105 rad/s²

Tangential acceleration a = r α = 6.2 x 0.105 = 0.651 m/s²

b)The maximum speed.

v = 2πr/T

= $\dfrac{2\pi \times 6.2}{1.5}$

= 25.97 m/s

So centripetal acceleration.

a = $\dfrac{v^2}{r}$

= $\dfrac{25.97^2}{6.2}$

= 108.781 m/s^2

= 11.1 g

in combination with the gravitation acceleration.

$a_{total} = \sqrt{(11.1g)^2+g^2}$

$a_{total}= 11.145 g$

6 0

4 years ago

A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕

Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0

3 years ago