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3 years ago

15

If element X has 40 protons will it also have 40 electrons

2 answers:

kkurt [141]3 years ago

5 0

That actually depends. Is it an atom or an ion? If it is a stable atom, it will have a neutral charge and an even number of protons and neutrons, however, noble gasses are the only independent single atoms that do not need to be bonded for this to occur in nature. Most elements are found as ions with uneven numbers of electrons and protons in order to become more stable.

Nataliya [291]3 years ago

4 0

Yes, atoms need to be equal in protons and electrons

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3 years ago

Read 2 more answers

Assume that at sea-level the air pressure is 1.0 atm and the air density is 1.3 kg/m3.

scoundrel [369]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

Pressure at sea level = 1 atm = 101300 Pa

we know

P = ρ g h

$h = \dfrac{P}{\rho\ g}$

$h = \dfrac{101300}{1.3\times 9.8}$

h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

at x = 0 air density = 0

at x= h ρ_l = ρ_sl

assuming density is zero at x - distance

$\rho_x = \dfrac{\rho_{sl}}{h}\times x$

now, Pressure at depth x

$dP = \rho_x g dx$

$dP = \dfrac{\rho_{sl}}{h}\times x g dx$

integrating both side

$P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx$

$P =\dfrac{\rho_{sl}\times g h}{2}$

now,

$h=\dfrac{2P}{\rho_{sl}\times g}$

$h=\dfrac{2\times 101300}{1.3\times 9.8}$

h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

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3 years ago

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3 years ago

You are designing a delivery ramp for crates containing exercise equipment. The 1890 N crates will move at 1.8 m/s at the top of

Mashcka [7]

Answer:

K = 588.3 N/m

Explanation:

From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):

Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:

$X = \frac{Ff + W*sin(22)}{K}=\frac{1223}{K}$

By conservation of energy:

$\frac{K*X^{2}}{2}-mg*d*sin(22)-\frac{m*V^{2}}{2}=-Ff*d$

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:

K = 588.3 N/m

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3 years ago