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likoan [24]
2 years ago
6

Mercury’s natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) c

ould mercury naturally exist?
Physics
1 answer:
Georgia [21]2 years ago
4 0

Answer:

Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?

Liquid is the answer

Explanation:

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The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance
belka [17]

The gravitational force between the spheres is

F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}

7 0
3 years ago
man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and h
-Dominant- [34]

Answer:

w₂ = 22.6 rad/s

Explanation:

This exercise the system is formed by platform, man and bricks; For this system, when the bricks are released, the forces are internal, so the kinetic moment is conserved.

Let's write the moment two moments

initial instant. Before releasing bricks

       L₀ = I₁ w₁

final moment. After releasing the bricks

       L_{f} = I₂W₂

       L₀ = L_{f}

       I₁ w₁ = I₂ w₂

       w₂ = I₁ / I₂ w₁

let's reduce the data to the SI system

     w₁ = 1.2 rev / s (2π rad / 1rev) = 7.54 rad / s

 

 let's calculate

       w₂ = 6.0/2.0   7.54

       w₂ = 22.6 rad/s

3 0
3 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

7 0
3 years ago
Why do electrical devices have resistance​
Kamila [148]
As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.
8 0
3 years ago
A. List at least 5 examples of events or factors that could cause extinction. 
baherus [9]
Disease
Lack of natural reourses
war
asteroid
Polution.

8 0
3 years ago
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