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3 years ago

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When heating a substance a phase change will start to occur when kinetic energy of the particles is _____ attractive forces bewt

ween the particles
low enough to overcome

great enough to combine

great enough to overcome

low enough to combine

equal to

1 answer:

Sergeu [11.5K]3 years ago

8 0

Great enough to overcome

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Need help ASAPPPPPP will mark brainleast if right no links or im reporting

kykrilka [37]

Answer: Density--> the amount of mass in a given volume

Boiling point: the tempeture when a liquid becomes gas

Conductivity: the ability of a subtance to transfer heat or electricety

Bouyant force: The upward force of an object in a liquid

soulibility: the ability of a subtance to disolve in other.

sorry for bad spelling

Explanation:

5 0

2 years ago

Read 2 more answers

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

3 years ago

How long does GSR last ?

SSSSS [86.1K]

Answer:

i think that the answer might be be C. one day. because it last about 4 to 6 hours

4 0

3 years ago

A certain piece of metal (density 9.45 grams per cubic centimeter) has the shape of a hockey puck with a diameter of 13 cm and a

atroni [7]

Answer:

0.0195 m

Explanation:

$\rho _{p}$ = density of hockey puck = 9.45 gcm⁻³ = 9450 kgm³

$d_{p}$ = diameter of hockey puck = 13 cm = 0.13 m

$h_{p}$ = height of hockey puck = 2.8 cm = 0.028 m

$\rho _{m}$ = density of mercury = 13.6 gcm⁻³ = 13600 kgm³

$d$ = depth of puck below surface of mercury

According to Archimedes principle, the weight of puck is balanced by the weight of mercury displaced by puck

Weight of mercury displaced = Weight of puck

$\rho _{m} (0.25)(\pi d_{p}^{2} d ) g = \rho _{p} (0.25)(\pi d_{p}^{2} h_{p} ) g\\\rho _{m} ( d ) = \rho _{p} ( h_{p} )\\(13600) d = (9450) (0.028)\\d = 0.0195 m$

7 0

3 years ago

Metal sphere having an excess of +5 elementary charges has a net electric charge of

Natasha2012 [34]

Answer:

$q=+8.01\cdot 10^{-19}\ coulombs$

Explanation:

<u>Elementary charge</u>

The elementary charge, denoted by the symbol e is the electric charge carried by a proton or, equivalently, the magnitude of a negative electric charge carried by an electron, which has charge −e.

The value of the elementary charge is a fundamental constant in physics:

$\mathbf{e}=1.60217662 \cdot 10^{-19}\ coulombs$

If a metal sphere has an excess of +5 elementary charge, then it has a net charge of:

$q=5*\mathbf{e}=+5*1.60217662 \cdot 10^{-19}\ coulombs$

$\boxed{q=+8.01\cdot 10^{-19}\ coulombs}$

5 0

3 years ago