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3 years ago

Allison has been diagnosed with interstitial cystitis. Which treatment is her doctor likely to recommend?

Physics

1 answer:

Basile [38]3 years ago

6 0

Answer: C) Sacral nerve stimulation

Explanation: interstitial cystitis can be simply refered to as a non infectious painful bladder condition which may be chronic depending on the severity. Some of the symptoms include pelvic and bladder pain as well as an urge to frequently urinate.

The best treatment that Allison's doctor is likely to recommend to her is the sacral nerve stimulation. Reducing the urgency to urinate which is associated with interstitial cystitis is the main target of this nerve stimulation technique, it involves simulating the sacral nerves which are the primary link between the spinal cord and nerves in the bladder. In this technique, electrical impulses are sent to the bladder by a thin wire which will be placed near the sacral nerves, this will help to reduce some of the symptoms.

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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago

The current atomic model has a ?

zavuch27 [327]

Answer:

Atoms have protons and neutrons in the center, making the nucleus, while the electrons orbit the nucleus. The modern atomic theory states that atoms of one element are the same, while atoms of different elements are different. What makes atoms of different elements different?

3 0

3 years ago

1 2 3 4 5 6 7 8 9 10

7nadin3 [17]

Answer:

<h2>C. 0.55 m/s towards the right</h2>

Explanation:

Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Momentum = Mass (M) * Velocity(V)

BEFORE COLLISION

Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s

Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s

AFTER COLLISION

Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s

x is the final velocity of the 0.25kg ball

Momentum of 0.15kg body moving at 0.75m/s(body at rest) =

0.15 * 0.75kgm/s = 0.1125 kgm/s

Using the law of conservation of momentum;

0.25+0 = 0.25x + 0.1125

0.25x = 0.25-0.1125

0.25x = 0.1375

x = 0.1375/0.25

x = 0.55m/s

Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at 0.55 m/s towards the right

4 0

3 years ago

A baseball has a mass of 0.145 kg. A professional pitcher throws a baseball 67 mi/h, which is 30.0 m/s. What is the magnitude of

Kamila [148]

The concept of momentum tells us that it is equivalent to the product between the mass and the velocity of the object, that is to say that in general it can be written as

$p = mv$