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DochEvi [55]
2 years ago
10

If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, what

does she have to do with the mass of the smiley face to keep the mobile in perfect balance? Note that she may have to change masses of other objects to keep the entire structure balanced
Physics
1 answer:
charle [14.2K]2 years ago
4 0

Answer:

She will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

Explanation:

mass of an object is directly proportional to the cube of its length. In this case the length is constant, the mass will also be constant for the smiley face, so that the mobile will be kept in perfect balance.

Therefore, If Anya decides to make the star twice as massive, and not change the length of any crossbar or the location of any object, she will make the mass of the smiley face twice as massive in order to keep the mobile in perfect balance.

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Bill has a mass of 85 kg and is skating west. He increases his speed from 3 m/s to 5 m/s by applying a force for 3 seconds. What
likoan [24]
F = ma
F = 85×(5-3)÷3
F = 85×(2÷3)
F = 85×0.667
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2 years ago
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}

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If a 1,300 kg car with no people inside is on the edge of a cliff 1,500 m above the ground, what is its potential energy?
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<u>Given that</u>

mass (m) = 1300 Kg ,

height (h) = 1500 m

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Answer:

1.a

2.longer

Explanation:

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