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3 years ago

5

__________ is a synthetic drug with effects very similar to those of cocaine. A. Ecstasy B. Dramamine C. Hydrocodone D. None of

the above

2 answers:

ehidna [41]3 years ago

4 0

Answer is C. Hydrocodone, also known as dihyrocodeinone, is a semi-synthetic of opioid synthesized from codeine, found in opium poppy. Used to relieve moderate to severe pain.

zhannawk [14.2K]3 years ago

3 0

C. hydrocodone is what i think.

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A 70.9-kg boy and a 43.2-kg girl, both wearing skates face each other at rest on a skating rink. The boy pushes the girl, sendin

Lelechka [254]

Answer:

Explanation:

Given

mass of boy $m_b=70.9\ kg$

mass of girl $m_g=43.2\ kg$

speed of girl after push $v_g=4.64\ m/s$

Suppose speed of boy after push is $v_b$

initially momentum of system is zero so final momentum is also zero because momentum is conserved

$P_i=P=f$

$0=m_b\cdot v_b+m_g\cdot v_g$

$v_b=-\frac{m_g}{m_b}\times v_g$

$v_b=-\frac{43.2}{70.9}\times 4.64$

$v_b=-2.82\ m/s$

i.e. velocity of boy is 2.82 m/s towards west

8 0

3 years ago

Dwayne ‘The Rock’ Johnson needs to escape from the fourth floor of a burning building (in a movie). He ties a rope around his wa

ZanzabumX [31]

Answer:

Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s

Explanation:

Let's start out with finding the force acting downwards because of the mass of 'The Rock':

Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N

Now the problem also states that the kinetic friction of the desk in this problem is 370 N

Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N

Now lets use F = ma to calculate for the acceleration of the desk:

787.58 = 63 x acceleration

acceleration = 12.501 m/s

Finally, we can use the motion equation:

$v^2 - u^2 = 2*a*s$

here u = 0 m/s (since initial speed of the desk is 0)

a = 12.501 m/s

and s = 10 m

Solving this we get:

$v^2 - 0 = 2 * 12.501 * 10$

$v = 15.812 m/s$

Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at 15.812 m/s when the desk reaches the window.

5 0

3 years ago

The Balmer series in hydrogen atom produces only infrared emission. O True O False

lorasvet [3.4K]

Answer:

False

Explanation:

As we know that, the Balmer series gives the n values as,

$n_{i}=2$ .

$[tex]n_{f}=3,4,5,.....\infty$ .

Now the value of wavelength can be calculated as,

$\frac{1}{\lambda}=R(\frac{1}{n_{i} }-\frac{1}{n_{f} } )z^{2}$ .

Here, $R=109677 cm^{-1}$ .

And $n_{f}=3$ .

Now,

$\frac{1}{\lambda}=109677 cm^{-1}(\frac{1}{2}-\frac{1}{3})$ .

Therefore,

$\lambda=\frac{6}{109677} cm\\\lambda=547\times 10^{-9} m\\ \lambda=547 nm$

Therefore, the wavelength of Balmer series lies in visible region which is 547 nm.

8 0

3 years ago

Read 2 more answers

An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia

Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

$e(t) = e* sin(wt)$

$e(t) = 6.04 * sin(Φ + π)$

$e(t) = 6.04 * sin(32.5 + 180)$

$e(t) = -3.245 V$

Now, Using Kirchoff Voltage Law,

$e(t) - VR- VL - VC = 0$

$-3.24 - 0 - VL - 5.32 = 0$

Finally we have the potential difference across the inductor.

$VL = - 8.56 v$

5 0

3 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N$

6 0

3 years ago