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insens350 [35]
3 years ago
13

True or False: The absorption of alcohol can be slowed down by eating, but only water can reduce the BAL level.

Physics
2 answers:
Andre45 [30]3 years ago
5 0

Answer:

False

Explanation:

The blood alcohol level cannot be changed by anything but the metabolism or the individual that is drinking the alcohol, and that is a job performed by the liver, so no one can speed up that process, alcohol absorption does change depending on what the person has in their stomach, that is why people get dunker mmore easily when they haven´t have anything to eat before drinking.

lesya [120]3 years ago
3 0
A hopes this helps idk this
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In the flow past a compression corner, the upstream Mach number and pressure are 3.5 and 1 atm, respectively. Downstream of the
yulyashka [42]

Answer:

\theta=23.7^{\circ}

Explanation:

The ratio of pressure 2 to 1 us 5.48/1= 5.48 rounded off as 5.5.

Referring to table A.2 of modern compressible flow then M_{\beta_1}=2.2

Also

M_{\beta_1}=M_1 sin \beta and making sin\beta the subject of the formula then

sin\beta=\frac {2.2}{3.5}\\\beta=38.94^{\circ}

Making reference to \theta-\beta-M diagram then

\theta=23.7^{\circ}

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A stone is thrown upward from the top of a building at an angle of 30° to the horizontal and with an initial speed of 20 m/s. Th
Hitman42 [59]

Answer:

Explanation:

Given

inclination \theta =30^{\circ}

initial speed u=20\ m/s

Point of release is 45 m above the ground

Considering stone to be a projectile, so time taken by projectile for its zero vertical displacement is

t_1=\frac{2u\sin \theta }{g}

t_1=\frac{2\times 20\times \sin 30}{10}

t_1=2\ s

Now after completing zero vertical displacement , stone needs to travel another 45 m in downward direction with initial speed u=20\sin 30

h=u_yt+\frac{1}{2}a_yt^2

where, h=height

u_y=vertical velocity

a_y=vertical acceleration

t_0=time

45=20\sin 30+\frac{1}{2}(9.8)(t_0)^2

t_0^2=\frac{70}{9.8}

t_0=2.64\ s

thus total time time required is t=t_0+t_1=2.64+2=4.64\ s

vertical velocity just before hitting

v_y=\sqrt{u_y^2+2\times a_y\times s}

v_y=\sqrt{10^2+2\times 10\times 45}

v_y=\sqrt{1000}=31.622\ m/s

Horizontal velocity v_x=u\cos 30=17.32\ m/s

Net velocity Just before hitting =\sqrt{v_x^2+v_y^2}

=\sqrt{(17.32)^2+(31.62)^2}

=\sqrt{1299.82}=36.05\ m/s

                 

7 0
3 years ago
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