All categories

2 years ago

Which of the following statements about blood alcohol concentration (BAC) are true?

Physics

2 answers:

mars1129 [50]2 years ago

5 0

D) Neither A or B are correct

KonstantinChe [14]2 years ago

3 0

Answer:

C: Both A and B are true

Explanation:

From DUI(Driving under influence) laws in most states in America, the DUI rule for underage individuals is that, it's illegal for an under 21 individual to have a blood alcohol content (BAC) of 0.08% or higher.

Now, in many states, it could a bit flexible with some considerations like, having the BAC to be as low as 0.02% or below or even having zero tolerance as the case may be.

Many of the states also have lose laws as well to this effect.

However, looking at the options, the correct ones will be options A & B because most states exhibit either options.

You might be interested in

A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At

Andreas93 [3]

The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

6 0

2 years ago

Julian and Joshua each maintain a constant speed as they run laps around a 400-meter track. In the time it takes Julian to compl

Marysya12 [62]

Answer:

the time Joshua travels 1 mile is 12.5 min

Explanation:

Let's start by finding the distance traveled on each lap,

Let's reduce everything to the SI system

R = 400 m

d = 1 mile (1609 m / 1 mile) = 1609 m

L = 2 pi R

L = 2 pi 400

L = 2513 m

Let us form a rule of proportions if 2 turns of Julian is 3 turns Joshua, for 1 turn of Joshua how many turns Julian took

lap Julian = 2/3 turn Joshua

Let's calculate what distance is the same for both of them since they are on the same track

1 lap = 2513 m

d. Julian = 2/3 2513 m

d Julian = 1675 m distance Joshua

Let us form the last rule of three or proportions if 1609 m you travel in 12 min how long it takes to travel 1675 m

t Julian = 1675/1609 12

t = 12.5 s

Since this is the distance Joshua travels, this is the time Joshua travels 1 mile

5 0

3 years ago

Read 2 more answers

Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+

babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• $\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j$ with $0\le x\le2$ and $0\le y\le6-x$ ;

• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

• $\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k$ with $0\le u\le\frac\pi2$ and $0\le y\le6-2\cos u$ .

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

$\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k$

$\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j$

$\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i$

$\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j$

$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0$

$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

$\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8$

$\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz$

$=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0$

$\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du$

$=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi$

$\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du$

$=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24$

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV$

where <em>R</em> is the interior of <em>S</em>. We have

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7$

The integral is easily computed in cylindrical coordinates:

$\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2$

$\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3$

as expected.

4 0

2 years ago

A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal

Lynna [10]

Wow ! This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

      (speed-squared) / (radius of the circle) .

Notice that we won't need to use the mass of the train.

We know the radius of the track. We don't know the trains speed yet,
but we do have enough information to figure it out. That's what we
need to do first.

Speed = (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) = 2.4π meters

10 laps = 24π meters.

Time = 1 minute 20 seconds = 80 seconds

The trains speed is (distance) / (time)

   = (24π meters) / (80 seconds)

   = 0.3 π meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

<span> (speed)² / (radius)

   = (0.3π m/s)² / (1.2 meters)

   = (0.09π m²/s²) / (1.2 meters)

   =    (0.09π / 1.2) m/s²

   =    0.236 m/s² . (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well, Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force. </span>

4 0

2 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

vladimir2022 [97]

The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
The force of friction, $F_f$ , acting in the backward direction

According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$