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3 years ago

Atoms are very small. Which measurement is the approximate diameter of a helium atom?

Physics

1 answer:

stich3 [128]3 years ago

3 0

Atoms diameter are on the order of 60 to 600pm (picometers). This is around 6*10^-11.

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A gas occupies a volume of 20 cubic meters at 9,000 pascals. If the pressure is lowered to 5,000 pascals, what volume will the g

forsale [732]

We need to consider no change in the temperature of gas (isothermal transformation)

Volume and pressure are inversely proportional magnitudes, so we can write:

$P_1.V_1=P_2.V_2\\
\\
9.20=5.V_2\\
\\
V_2=\frac{180}{5}=36 \ m^3$

5 0

2 years ago

In 2020, nasa confirmed the existence of what on the lunar surface?.

kipiarov [429]

Water was confirmed to be on the sunlit surface of the Moon

6 0

2 years ago

Can anyone help me on the summer and winter Gizmo questions. Activities A and B?

Anuta_ua [19.1K]

Answer:

sure, do you have more detailed of the assignments...?

Explanation:

6 0

2 years ago

An underground gasoline tank can hold 1.07 103 gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outd

Damm [24]

Answer:

1069.38 gallons

Explanation:

Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.

V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹

Δθ = (5/9)(97°F -52°F) °C = 25 °C.

Let V₂ be its final volume when it cools to 52°F in the tank is

V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)

= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)

= 1.07 × 10³(1 - [0.024]²)

= 1.07 × 10³(1 - 0.000576)

= 1.07 × 10³(0.999424)

= 1069.38 gallons

7 0

3 years ago

Air flows through an adiabatic turbine that is in steady operation. The air enters at 150 psia, 900oF, and 350 ft/s and leaves a

Nonamiya [84]

Answer:

$1486.5\frac{Btu}{s}$

Explanation:

The inlet specific volume of air is given by:

$v_1=\frac{RT_1}{P_1}\\\\v_1=\frac{(0.3704\frac{psia.ft^3}{lbm.R})(1360R)}{150psia}\\\\v_1=3.358\frac{ft^3}{lbm} \ \ \ \ \ \ \ \ \...i$

The mass flow rates is expressed as:

$\dot m=\frac{1}{v_1}A_1V_1\\\\\dot m=\frac{1}{3.358ft^3/psia}(0.1ft^2)(350ft/s)\\\\\dot m=10.42\frac{lbm}{s}$

The energy balance for the system can the be expresses in the rate form as:

$E_{in}-E_{out}=\bigtriangleup \dot E=0\\\\E_{in}=E_{out}\\\\\dot m(h_1+0.5V_1^2)=\dot W_{out}+\dot m(h_2+0.5V_2^2)+Q_{out}\\\\\dot W_{out}=\dot m(h_2-h_1+0.5(V_2^2-V_1^2))=-m({cp(T_2-t_1)+0.5(V_2^2-V_1^2)})\\\\\\\dot W_{out}=-(10.42lbm/s)[(0.25\frac{Btu}{lbm.\textdegree F})(300-900)\textdegree F+0.5((700ft/s)^2-(350ft/s)^2)(\frac{1\frac{Btu}{lbm}}{25037ft^2/s^2})]\\\\\\\\=1486.5\frac{Btu}{s}$

Hence, the mass flow rate of the air is 1486.5Btu/s

5 0

2 years ago