A substance that is more sense has more mass per unit volume, it will make sound to transmit at slower rate . Less dense substance will allow sound to transmit at higher rates. Density affect the speed a wave will be transmitted through it. The denser the material, the slower wave will be transmitted.

Explanation:

Density is Mass per unit volume. Density affect the propagation of wave. The denser the substance, the slower wave will be transmitted through it. The less dense the material, the higher wave will be transmitted through it. Density affect the speed of wave.

8 0

3 years ago

Instill, in medical terms, generally refers to Question 20 options: a) injecting into veins. b) placing medication drops in nose

dolphi86 [110]

Answer:

The correct answer is b) placing medication drops in nose, eyes, or ears.

Explanation:

Instill refers to the action of pouring or introducing a liquid drop by drop on a mucosal surface or inside a canal or organic cavity, for therapeutic purposes. The most used instillations are: conjunctival, nasal, paranasal, atrial, laryngeal, tracheal, bronchial, bladder and urethral. The instillation can be carried out with the help of a simple common dropper or with an instrument called an instillator, which consists of a probe with a widened end and a syringe with a screw plunger; by activating the screw, the medicine is introduced drop by drop.

7 0

3 years ago

A fossil is discovered that has only 12.5% of the carbon-14 that it would have had originally (when the animal was alive). If th

mariarad [96]

Answer:

The fossil is 17,100 years old.

Explanation:

The decay equation:

$\frac{dN}{dt}\propto -N$

$\Rightarrow \frac{dN}{dt}= -\lambda N$

$\Rightarrow \frac{dN}{N}= -\lambda dt$

Integrating both sides

$\Rightarrow \int\frac{dN}{N}= \int-\lambda dt$

$\Rightarrow ln |N|=-\lambda t+c$

When t=0, N= $N_0$ = initial amount

$ln |N_0|=-\lambda .0+c$

$\Rightarrow ln |N_0|=c$

$\therefore ln |N|=-\lambda t+ln|N_0|$

$\Rightarrow ln |N|-ln|N_0|=-\lambda t$

$\Rightarrow ln |\frac {N}{N_0}|=-\lambda t$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow N=N_0e^{-\lambda t}$

The decay equation is

$N=N_0e^{-\lambda t}$

Given that,

The half life of carbon - 14 is 5700 years.

For half life, $N=\frac{1}{2} N_0$

To find the value of $\lambda$ , we need to put the value of N and t in the decay equation.

$\frac12N_0=N_0e^{-\lambda \times 5700}$

$\Rightarrow \frac12=e^{-\lambda \times 5700}$ [ Divided $N_0$ both sides]

Taking ln both sides

$\Rightarrow ln| \frac12|=ln|e^{-\lambda \times 5700}|$

$\Rightarrow ln| \frac12|={-\lambda \times 5700}$

$\Rightarrow \lambda= \frac{ln| \frac12|}{-5700}$

$\Rightarrow \lambda= \frac{ln|1|-ln|2|}{-5700}$ [ $ln|\frac mn|= ln |m|-ln |n|$ ]

$\Rightarrow \lambda= \frac{ln|2|}{5700}$ [ln 1= 0]

The fossil has only 12.5% of the carbon carbon-14 that it would have had originally.

So, $N=\frac{12.5}{100} N_0$

Then,

$\frac{12.5}{100} N_0=N_0e^{-\frac{ln|2|}{5700}t$

$\Rightarrow \frac{12.5}{100} =e^{-\frac{ln|2|}{5700}t$

Taking ln both sides

$\Rightarrow ln|\frac{12.5}{100} |=ln|e^{-\frac{ln|2|}{5700}t}|$

$\Rightarrow ln|\frac{12.5}{100} |={-\frac{ln|2|}{5700}t}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|} {-\frac{ln|2|}{5700}}$

$\Rightarrow t=\frac{ ln|\frac{12.5}{100}|\times 5700} {-{ln|2|}}$

$\Rightarrow t=17,100$

The fossil is 17,100 years old.

7 0

4 years ago

In your own words, outline the sequence of steps involved in the process of going from the DNA base pairs of TAC to the amino ac

sveticcg [70]

The first stage of gene expression is the transcription step, ie the passage from ADn to mRNA, the RNA polymerase will read DNA (the TAC sequence) and will synthesize a complementary strand (AUG) in the form of mRNA (which will then undergo mRNA processing).

The second stage of gene expression is the translation of mRNA into protein: the AUG sequence of mRNA (supposed to be the start codon) is translated to the amino acid, methionine, via the ribosome and tRNA. The ribisome will read adjacent codons to form the polypeptide (several amino acids bonded together by peptide bonds).

3 0

4 years ago