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2 years ago

Which of the following is an example of an analgesic

Physics

2 answers:

ludmilkaskok [199]2 years ago

7 0

Answer:

since the choices to pick from were not provided, I have decided to give a list of analgesic examples.

analgesic combinations

antimigraine agents

CGRP inhibitors

cox-2 inhibitors

miscellaneous analgesics

narcotic analgesic combinations

narcotic analgesics

Nonsteroidal anti-inflammatory drugs

salicylates

Nookie1986 [14]2 years ago

5 0

Answer:

Aspirin

Explanation:

Aspirin is an effective analgesic for acute pain, although it is generally considered inferior to ibuprofen because aspirin is more likely to cause gastrointestinal bleeding. Aspirin is generally ineffective for those pains caused by muscle cramps, bloating, gastric distension, or acute skin irritation.

Other name: 2-acetoxybenzoic acid

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Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350°F, releasing a wonderful aroma

Morgarella [4.7K]

Answer:

The mass of the baked loaf will be less than the dough.

Explanation: When heat is applied to food substance or products like the one pound the substance or material gains a higher temperature, the increase in temperature causes moisture inherent or added to the product in this case the one pound dough to be lost, the one pound dough prepared at room temperature, once it is placed inside the oven at 350 degrees Fahrenheit it will lose moisture in the form of vapor to the environment as noticed in the aroma, the moisture lost will eventually reduce it mass/weight (kilograms or grams) by some percentage or quantities(kilograms or grams)

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3 years ago

A varying force is given by F=Ae ^-kx, where x is the position;A and I are constants that have units of N and m^-1 , respectivel

Burka [1]

W = ∫ (x from 0.1 to +oo) F dx

= ∫ (x from 0.1 to +oo) A e^(-kx) dx

= A/k x [ - e^(-kx) ](between 0.1 and +oo)

= A/k x [ 0 + e^(-k * 0.1) ]

<span> = A/k x e^(-k/10) </span>

4 0

3 years ago

The current in a single-loop circuit with one resistance R is 6.3 A. When an additional resistance of 3.4 Ω is inserted in serie

Dmitry [639]

Answer:

10.15Ω

Explanation:

From ohm's law,

V = IR...................... Equation 1

Where V = Voltage, I = current, R = resistance.

Assume the voltage across the resistance = V,

Given: I = 6.3 A

Substitute into equation 1

V = 6.3R.................. Equation 2

When an additional resistance of 3.4 Ω is inserted in series with R,

The voltage remain the same, but the current changes

Total Resistance(Rt) = (R+3.4)Ω, I' = 4.72 A

Also from ohm' law,

V = I'Rt............... Equation 3

Substitute the value of I' and Rt into equation 3

V = 4.72(R+3.4)............... Equation 5.

Divide equation 2 by equation 5

V/V = 6.3R/4.72(R+3.4)

1 = 1.335R/(R+3.4)

R+3.4 = 1.335R

3.4 = 1.335R-R

3.4 = 0.335R

R = 3.4/0.335

R = 10.15Ω

6 0

2 years ago

Read 2 more answers

The floor of a railroad flatcar is loaded with loose crates having a coefficient of static friction of 0.32 with the floor. If t

coldgirl [10]

Answer:

The shortest braking distance is 35.8 m

Explanation:

To solve this problem we must use Newton's second law applied to the boxes, on the vertical axis we have the norm up and the weight vertically down

On the horizontal axis we fear the force of friction (fr) that opposes the movement and acceleration of the train, write the equation for each axis

Y axis

N- W = 0

N = W = mg

X axis

-Fr = m a

-μ N = m a

-μ mg = ma

a = μ g

a = - 0.32 9.8

a = - 3.14 m/s²

We calculate the distance using the kinematics equations

Vf² = Vo² + 2 a x

x = (Vf² - Vo²) / 2 a

When the train stops the speed is zero (Vf = 0)

Vo = 54 km/h (1000m/1km) (1 h/3600s)= 15 m/s

x = ( 0 - 15²) / 2 (-3.14)

x= 35.8 m

The shortest braking distance is 35.8 m

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2 years ago

A hot air balloon is on the ground, 200 feet from an observer. The pilot decides to ascend at 100 ft/min. How fast is the angle

liq [111]

Answer:

0.0031792338 rad/s

Explanation:

$\theta$ = Angle of elevation

y = Height of balloon

Using trigonometry

$tan\theta=y\dfrac{y}{200}\\\Rightarrow y=200tan\theta$

Differentiating with respect to t we get

$\dfrac{dy}{dt}=\dfrac{d}{dt}200tan\theta\\\Rightarrow \dfrac{dy}{dt}=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow 100=200sec^2\theta\dfrac{d\theta}{dt}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{100}{200sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{1}{2}cos^2\theta$

Now, with the base at 200 ft and height at 2500 ft

The hypotenuse is

$h=\sqrt{200^2+2500^2}\\\Rightarrow h=2507.98\ ft$

Now y = 2500 ft

$cos\theta=\dfrac{200}{h}\\\Rightarrow cos\theta=\dfrac{200}{2507.98}=0.07974$

$\dfrac{d\theta}{dt}=\dfrac{1}{2}\times 0.07974^2\\\Rightarrow \dfrac{d\theta}{dt}=0.0031792338\ rad/s$

The angle is changing at 0.0031792338 rad/s

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3 years ago