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2 years ago

A patient who smokes has hypertension. how could smoking have contributed to the patients hypertension

2 answers:

6 0

The nicotine in cigarette smoke is a big part of hypertension. It raises your blood pressure and heart rate, makes your arteries more narrow and hardens their walls, and also makes your blood more likely to clot. Hope this answers the question.

Sergio [31]2 years ago

5 0

By constricting blood vessels

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What voltage would produce a current of 100amps through an aluminum wire assuming that the re sistance of the wire is3.44×10-4 o

agasfer [191]

V = IR

I = current
R = resistance

Voltage = 100 * (3.44x 10^-4) = do the calculation

Hope this helps

8 0

2 years ago

Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe

uranmaximum [27]

Answer:

a) 105935.7 Pa

b) 103630.35 Pa

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = $\pi r^2h$

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x $0.13^2$ x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = 105935.7 Pa

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = 103630.35 Pa

7 0

3 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

3 0

3 years ago

Most clouds form in the atmosphere when moist air 1. rises,expands,and cools to the dewpoint 2. rises, expands, and warms to the

earnstyle [38]

Most clouds form in the atmosphere when moist air rises expands and cool to the dew point. When the air above reaches its saturation point, the water vapor is attracted to dust particles which when they accumulate the hold on tho each other and form clouds. Evaporation and condensation is what causes saturation above. When the clouds become heavy enough with moisture, the water then fall to earth as rain.

4 0

3 years ago

the resistance of a wire of length 80cm and of uniform area of cross-section 0.025cmsq., is found to be 1.50 ohm. Calculate spec

vivado [14]

$Specific\ resistance\ =resistivity\\
From\ formula\ on \ resistance:\ R= \frac{pL}{A}\ p-resistivity,\ L-length,\\ A-area\ of\ cross\ section\\
p= \frac{R*A}{L}= \frac{1,5Ohm*0,025*10^{-4}m^2 }{80* 10^{-2}m }=0,0003515625 Ohm*m$

3 0

3 years ago