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1 year ago

Who pays for Government workers that work on alcohol impaired driving cases?

Physics

1 answer:

Shalnov [3]1 year ago

8 0

Answer: Taxpayers

Explanation:

Taking alcohol before driving or while driving is dangerous and has resulted in lots of accidents and deaths. Alcohol tampers with the normal functioning of the brain, and also impairs ones reasoning.

Alcohol impaired driving cases handled by government officials are paid for by the taxpayers. A tax is the levy that the people in the country pays. Those funds are used in handling different government objectives and this is one of such ways.

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Moving water, like that of a river, carries sediment as it moves along its bed. The faster the water flows, the more sediment th

katovenus [111]

Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

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1 year ago

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If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

zalisa [80]

If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.

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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

Consider a vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + 1 4 u' + 2

GarryVolchara [31]

Answer:

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

Explanation:

Given vibrating system is

$u''+\frac{1}{4}u'+2u= 2cos \omega t$

Consider U(t) = A cosωt + B sinωt

Differentiating with respect to t

U'(t)= - A ω sinωt +B ω cos ωt

Again differentiating with respect to t

U''(t) = - A ω² cosωt -B ω² sin ωt

Putting this in given equation

$-A\omega^2cos\omega t-B\omega^2sin \omega t+ \frac{1}{4}(-A\omega sin \omega t+B\omega cos \omega t)+2Acos\omega t+2Bsin\omega t = 2cos\omega t$

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)cos \omega t+(-B\omega^2-\frac{1}{4}A\omega+2B)sin \omega t= 2cos \omega t$

Equating the coefficient of sinωt and cos ωt

$\Rightarrow (-A\omega^2+\frac{1}{4}B\omega +2A)= 2$

$\Rightarrow (2-\omega^2)A+\frac{1}{4}B\omega -2=0$ .........(1)

and

$\Rightarrow -B\omega^2-\frac{1}{4}A\omega+2B= 0$

$\Rightarrow -\frac{1}{4}A\omega+(2-\omega^2)B= 0$ ........(2)

Solving equation (1) and (2) by cross multiplication method

$\frac{A}{\frac{1}{4}\omega.0 -(-2)(2-\omega^2)}=\frac{B}{-\frac{1}{4}\omega.(-2)-0.(2-\omega^2)}=\frac{1}{(2-\omega^2)^2-(-\frac{1}{4}\omega)(\frac{1}{4}\omega)}$

$\Rightarrow \frac{A}{2(2-\omega^2)}=\frac{B}{\frac{1}{2}\omega}=\frac{1}{(2-\omega^2)^2+\frac{1}{16}\omega}$

$\therefore A=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega}$ and $B=\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega}$

Therefore the required solution is

$U(t)=\frac{2(2-\omega^2)^2}{(2-\omega^2)^2+\frac{1}{16}\omega} cos\omega t +\frac{\frac{1}{2}\omega}{(2-\omega^2)^2+\frac{1}{16}\omega} sin \omega t$

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2 years ago

In an experiment, Mary compares a collision of two 1-kilogram steel balls with a collision of two 1-kilogram foam rubber balls.

IgorC [24]

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1 year ago

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