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2 years ago

9

A floating balloon can be formed when the substance helium is released from a compressed container into a flat rubber balloon. T

hen, the decompressed helium atoms expand and float up, making the rubber balloon expand around them and float with them. What is helium’s state when it does this? liquid solid gas plasma

2 answers:

mel-nik [20]2 years ago

6 0

Helium’s state when the decompressed helium atoms expand and float up, making the rubber balloon expand around them and float with them is gas.

<span>Helium is a chemical element with symbol He and atomic number 2. It is a colorless, odorless, tasteless, non-toxic, inert, monatomic gas, the first in the noble gas group in the periodic table. Its boiling point is the lowest among all the elements.</span>

Delicious77 [7]2 years ago

3 0

<h3><u>Answer;</u></h3>

Gas

<h3><u>Explanation;</u></h3>

The helium is in gaseous state.
Helium is an element that exists in gaseous state, it belongs to group 8 in the periodic table called noble gases.
A gas balloon is a balloon which flies since it is filled with a gas which is less dense than air or lighter than air. Such gases include; helium gas and hydrogen gas.

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The velocity of a car increases from 2.0 m/s to 16.0 m/s in a time period of 3.5 s. What was the average acceleration?

dangina [55]

Answer:

the acceleration is a=3ms^-2

Explanation:

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2 years ago

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100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

8 0

2 years ago

A pump moves water horizontally at a rate of 0.02 m3/s. Upstream of the pump where the pipe diameter is 90 mm, the pressure is 1

victus00 [196]

Answer:

the efficiency of hydralic is 79.88%

Explanation:

convert mm to m

1mm = (1/1000)m

diameter of pipe upsteam

d₁= 90mm= 0.09m

diameter of pipe downsteam

d₂= 30mm = 0.03m

finding velocity of upsteam

recall Q=A₁V₁

V₁=Q/A₁

V₁=3.14m/s

velocity of downsteam

V₂= Q/A₂

V₂= 28.29m/s

mass flow rate

m= ρQ

ρ is the density of water

m = 1000× 0.02

m= 20kg/s

the efficiency of hydralic is 79.88%

6 0

2 years ago

g A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobb

Law Incorporation [45]

Answer:

Explanation:

total weight acting downwards

= 3g + 10g

13 g

volume of lead = 10 / 11.3 = .885 cm³

Let the volume of bobber submerged in water be v in floating position . buoyant force on bobber = v x 1 x g

Buoyant force on lead = .885 x 1 x g

total buoyant force = vg + .885 g

For floating

vg + .885 g = 13 g

v = 12.115 cm³

total volume of bobber

= 4/3 x 3.14 x 2³

= 33.5 cm³

fraction of volume submerged

= 12.115 / 33.5

= .36

= 36 %

4 0

2 years ago

Driving your Ferrari through the Italian countryside at a speedy 88 m/s, you approach an opera diva singing a high C (1,046 Hz).

MrRissso [65]

Answer:

You will hear the note E₆

Explanation:

We know that:

Your speed = 88m/s

Original frequency = 1,046 Hz

Sound speed = 340 m/s

The Doppler effect says that:

$f' = \frac{v \pm v0 }{v \mp vs}*f$

Where:

f = original frequency

f' = new frequency

v = velocity of the sound wave

v0 = your velocity

vs = velocity of the source, in this case, the source is the diva, we assume that she does not move, so vs = 0.

Replacing the values that we know in the equation we have:

$f' = \frac{340 m/s + 88m/s}{340 m/s} *1,046 Hz = 1,316.73 Hz$

This frequency is close to the note E₆ (1,318.5 Hz)

7 0

1 year ago