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3 years ago

What is one key physical difference between transition metals and poor metals?

Physics

1 answer:

adoni [48]3 years ago

5 0

Answer:

HARDNESS

Explanation:

One key physical difference between transition metals and poor metals is their "" Hardness"" which is

the ability of a material to resist deformation. The test for hardness can be determined by a standard test which is the measurement of surface resistance to indentation. hardness tests are defined the shape and also type of indent.

The poor metals are also referred to as post transition metals. They are elements that are found at the right of the transition metals,they are located in the p-block,Their properties is as a result of their low melting and boiling point compare to other metals.They have high electronegativity and conductivity but softer texture compare to other metals.They are very soft more than the transition metals, but they cannot be cannot be classified as metalloids.

Poor metals includes elements in the periodic table such as; aluminium, gallium, indium, thallium, tin, lead, bismuth, and polonium.

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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

Kinetic energy of the tennis ball, KE' = $\frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s$

Now, the distance the ball can penetrate to is given by:

$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

$\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js$

Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

$P(t) = e^{\frac{- 2t}{\eta}}$

$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

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3 years ago

Which statement correctly describes the relationship between frequency and wavelength?

Len [333]

The relationship between the frequency and wavelength of a wave is given by the equation:

v=λf, where v is the velocity of the wave, λ is the wavelength and f is the frequency.

If we divide the equation by f we get:

λ=v/f

From here we see that the wavelength and frequency are inversely proportional. So as the frequency increases the wavelength decreases.

So the second statement is true: As the frequency of a wave increases, the shorter the wavelength is.

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3 years ago

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If it is fixed at C and subjected to the horizontal 60-lblb force acting on the handle of the pipe wrench at its end, determine

pickupchik [31]

Answer:

τ = 132.773 lb/in² = 132.773 psi

Explanation:

b = 12 in

F = 60 lb

D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in

d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in

We can see the pic shown in order to understand the question.

Then we get

Mt = b*F*Sin 30°

⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in

Now we find ωt as follows

ωt = π*(R⁴ - r⁴)/(2R)

⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)

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then the principal stresses in the pipe at point A is

τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)

⇒ τ = 132.773 lb/in² = 132.773 psi

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