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Ne4ueva [31]
4 years ago
12

True or False: If necessary, your blood may be drawn in DUI cases involving serious bodily injury or death by authorized medical

personnel with the use of reasonable force by the arresting officer, even if you, as the driver, refuse.
Physics
2 answers:
Eduardwww [97]4 years ago
8 0

True because sometimes a breathalyzer isn't as accurate as we think and if there is multiple sources of blood they would want to rule out who's blood is who's.
larisa [96]4 years ago
3 0
True without a doubt blood tests for alcohol and drugs is the main use
You might be interested in
Show explicitly that ▽ . B-0 near a long straight wire carrying a current I. HINT: you may use Cartesian coordinates, but anothe
aivan3 [116]

Answer:

\bigtriangledown.B=0 is proved.

Explanation:

The magnetic field in the long current carrying wire is,

B=\frac{\mu_{0}I }{2\pi r } \phi

Here, I is the current, B is the magnetic field.

Now, by using cylindrical coordinates for the divergence of B.

\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi} B

Put the value of B in above equation.

\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi}(\frac{\mu_{0}I }{2\pi r } \phi)\\\bigtriangledown.B=0

Hence, it is prove that for a long current I carrying wire magnetic field divergence that is \bigtriangledown.B=0.

7 0
3 years ago
This question is about the alpha particle experiment
Wewaii [24]
This suggest the the atom has a very small positively charged nucleus in the mass of the atom is concentrated. 
And the electrons revolves around the nucleus in their orbits.
4 0
4 years ago
If the magnitude of the moment of F about line CD is 57 N·m, determine the magnitude of F.If the magnitude of the moment of F ab
bazaltina [42]

Answer:

F_ab = 260.17 N

Explanation:

Given:

- Moment of force F about CD, (M)_cd = 57 Nm

Find:

- First we will write down the position vectors of points A, B , C , D:

- We will take left and bottom most corner of cube to be the origin.

- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.

- The position vectors wrt to the origin are:

                             Point A = 0.2 k

                             Point B = 0.4 i + 0.2 j

                             Point C = 0.2 j + 0.4 k

                             Point D = 0.4 i + 0.4 k

- Now we will determine the Force vector F_ab along vector AB.

                             vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k

                             unit (AB) = 0.4 i + 0.2 j - 0.2 k  / sqrt ( 0.4^2 + 2*0.2^2)

                                            = [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

Hence,

                              vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )

- Now, form a unit vector along the line CD:

                             vec(CD) = D - C = 0.4 i - 0.2 j

                             unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)

                                           = [sqrt(5)]*(0.4 i - 0.2 j)

- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:

                               vec(AC) =r_ac = C - A = 0.2 j + 0.2 k

- Now the moment about a line CD due to force is:

                              (M)_cd = unit(CD) . ( r_ac x vec(F_ab) )

The cross product of r_ac and vec(F_ab) is as follows:

                               (M)_c =  ( r_ac x vec(F_ab) ) :

                                \left[\begin{array}{ccc}i&j&k\\0&0.2&0.2\\0.8165&0.40824&-0.40824\end{array}\right]

                              (M)_c =  F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]

The dot product of (M)_c and unit (CD)  is as follows:

                              (M)_cd = unit(CD) . (M)_c :

 (M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] .  [sqrt(5)]*(0.4 i - 0.2 j)

                              (M)_cd = F_ab*(sqrt(30) / 25)

- The given magnitude of the moment is (M)_cd. Calculate F_ab:

                               57 = F_ab*(sqrt(30) / 25)  

                              F_ab = 260.17 N

7 0
4 years ago
An object is thrown downward with an ini-
aniked [119]

Answer:

69

Explanation:

8 0
4 years ago
a 10 kg block is attached to a light chord that is wrapped around the pulley of an electric motor. at what rate is the motor doi
defon

Answer:

The rate of work is 360 W

Explanation:

Given:

Speed of block v = 3 \frac{m}{s}

Upward acceleration of block a = 2 \frac{m}{s^{2} }

Mass of block m = 10 kg

First Find the force act on block due to gravity

  F = M (a + g)

Where g = 10 \frac{m}{s^{2} }

  F = 10 \times (10+2)

  F = 120 N

For finding at what rate motor doing work when it is pulling the block upward,

  P = F.v

  P = 120 \times 3

  P = 360 W

Therefore, the rate of work is 360 W

6 0
4 years ago
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