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3 years ago

Part B

Physics

1 answer:

erik [133]3 years ago

5 0

Answer:

Sound waves in liquids and gases involve alternating compression and rarefaction of material along a line defining the direction of propagation of the wave. These waves are known as longitudinal waves, and of course exist only in a medium that can be compressed and rarefied. In solids, sound energy also produces longitudinal waves, but it can also produce transverse waves, in which compression and rarefaction occurs perpendicular to the direction of propagation. These two waves propagate at different speeds, a phenomenon that is most noticeable in earthquakes. The first wave gives notice that the quake is coming, the second one does the damage. The time between the two tells you how far away the epicenter is. In water there is another kind of wave, called a gravity wave, the kind you see at the beach. All of these wave require a medium. There is no sound in a vacuum.

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A client with hypertension who weighs 72.4 kg is receiving an infusion of nitroprusside (Nipride) 50 mg in D5W 250 ml at 75 ml/h

Mkey [24]

To solve this problem it is necessary to simply apply the concepts related to cross-multiply and proportion between units.

Let's start first by relating the amount of dose needed to be supplied per hour, in other words,

The infusion of 250ml should be supplied at a rate of 75ml / hour, so what amount x of mg hour should be supplied with 50Mg.

$\frac{x}{75ml/hour} \rightarrow \frac{50mg}{250ml}$

$x \rightarrow \frac{50mg*75ml/hour}{250ml}$

$x \rightarrow \frac{3750mg}{250hour}$

$x \rightarrow 15\frac{mg}{hour}$

Converting to mcg units we know that 1mg is equal to 1000mcg and that 1 hour contains 60 min, therefore

$x \rightarrow 15\frac{mg}{hour}$

$x \rightarrow 15\frac{mg}{hour}(\frac{1000mcg}{1mg})(\frac{1hour}{60min})$

$x \rightarrow 250mcg/min$

The dose should be distributed per kilogram of the patient so if the patient weighs 72.4kg,

$Dose = \frac{250mcg/min}{72.4kg}$

$Dose = 3.5 \frac{mcg/min}{kg}$

Therefore the client will receive 3.5mcg/kg/min.

8 0

3 years ago

18. Un avión de rescate de animales que vuela hacia el este a 36.0 m/s deja caer una paca de

Alona [7]

Answer:

Definimos momento como el producto entre la masa y la velocidad

P = m*v

(tener en cuenta que la velocidad es un vector, por lo que el momento también será un vector)

Sabemos que el peso de la paca de heno es 175N, y el peso es masa por aceleración gravitatoria, entonces.

Peso = m*9.8m/s^2 = 175N

m = (175N)/(9.8m/s^2) = 17.9 kg

Ahora debemos calcular la velocidad de la paca justo antes de tocar el suelo.

Sabemos que la velocidad horizontal será la misma que tenía el avión, que es:

Vx = 36m/s

Mientras que para la velocidad vertical, usamos la conservación de la energía:

E = U + K

Apenas se suelta la caja, esta tiene velocidad cero, entonces su energía cinética será cero y la caja solo tendrá energía potencial (Si bien la caja tiene velocidad horizontal en este punto, por la superposición lineal podemos separar el problema en un caso horizontal y en un caso vertical, y en el caso vertical no hay velocidad inicial)

Entonces al principio solo hay energía potencial:

U = m*g*h

donde:

m = masa

g = aceleración gravitatoria

h = altura

Sabemos que la altura inicial es 60m, entonces la energía potencial es:

U = 175N*60m = 10,500 N

Cuando la paca esta próxima a golpear el suelo, la altura h tiende a cero, por lo que la energía potencial se hace cero, y en este punto solo tendremos energía cinética, entonces:

10,500N = (m/2)*v^2

De acá podemos despejar la velocidad vertical justo antes de golpear el suelo.

√(10,500N*(2/ 17.9 kg)) = 34.25 m/s

La velocidad vertical es 34.25 m/s

Entonces el vector velocidad se podrá escribir como:

V = (36 m/s, -34.25 m/s)

Donde el signo menos en la velocidad vertical es porque la velocidad vertical es hacia abajo.

Reemplazando esto en la ecuación del momento obtenemos:

P = 17.9kg*(36 m/s, -34.25 m/s)

P = (644.4 N, -613.075 N)

6 0

3 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

What is the percentage of the incoming solar radiation from the top of the atmosphere the gets reflected right back out into spa

san4es73 [151]

Answer: We can define the solar constant as a measure of the luminous flux density.

Explanation:

The solar constant or solar constant is the amount of energy radiated at the upper limit of the Earth's atmosphere per unit time perpendicular to the unit surface, at the Earth's mean distance from the sun. Amounts to 1367.7 W / m² ± 6 W / m². The sun's constant includes all kinds of electromagnetic radiation, not just visible light. The average value is 1,368 kW / m2 and changes slightly with solar cycles. The amount of these constant changes over one year and has different benefits.

4 0

3 years ago

Tarzan swings on a 31.0 m long vine initially inclined at an angle of 36.0◦ with the vertical. The acceleration of gravity if 9.

Gennadij [26K]

Answer:

$v=10.777m/s$

Explanation:

Tarzan swing can be thought of as change in potential energy by going from higher location We solve for height of beginning of the swing by using simple cosine equation:

$31Cos36=25.08\\E_{potential}=mgh\\$

ΔE=mg(h₂-h₁)

$=m*9.81(31-25.08)\\$

The potential energy of Tarzan initial position is converted into kinetic energy of his swing.By using kinetic equation

$E_{kinectic}=P_{potential}\\1/2mv^{2}=m*9.81(31.0-25.08)\\(1/2)v^{2}=9.81(31.0-25.08)\\0.5v^{2}=58.07\\v^{2}=58.07/0.5\\v=\sqrt{58.07/0.5}\\ v=10.777m/s$

8 0

3 years ago