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alexandr1967 [171]
3 years ago
5

In 1896, S. Riva-Rocci developed the prototype of the current sphygmomanometer, a device used to measure blood pressure. When it

was worn as a cuff around the upper arm and inflated, the air pressure within the cuff was connected to a mercury manometer. The density of mercury is 13,550 kg/m3. Assume all pressures are gage pressures Part A The reading for the high (or systolic) pressure is 120 mm. Determine this pressure in pascals and psi. Express your answers using three significant figures separated by a comma. pPa, ppsi = nothing Pa, psi Request Answer Part B The reading for the low (or diastolic) pressure is 80 mm. Determine this pressure in pascals and psi. Express your answers using three significant figures separated by a comma. pPa, ppsi = nothing Pa, psi
Physics
1 answer:
Anastasy [175]3 years ago
4 0

Answer:

Explanation:

A pressure that causes the Hg column to rise 1 millimeter is called a torr. The term 1 mmHg used can replaced by the torr.

1 atm = 760 torr = 14.7 psi.

A.

120 mmHg

Psi:

760 mmHg = 14.7 psi

120 mmHg = 14.7/760 * 120

= 2.32 psi

Pa:

1mmHg = 133.322 Pa

120 mmHg = 120 * 133.322

= 15998.4 Pa

B.

80 mmHg

Psi:

760 mmHg = 14.7 psi

80 mmHg = 14.7/760 * 80

= 1.55 psi

Pa:

1mmHg = 133.322 Pa

80 mmHg = 80 * 133.322

= 10665.6 Pa

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Calculate the root mean square velocity of nitrogen molecules at 25°c.
umka2103 [35]

Answer:

515.22 m/s

Explanation:

Formula for root mean square velocity is SQRT (3RT/M)

Where,

R = Rydberg's constant = 8.314 Kg. m²/ s². K. mol

T = Temperature in Kelvin

M = Molar mass in Kg/mol

First calculate molar mass,

As nitrogen is diatomic molecule,

Its molar mass is

M = 2(14) = 28 g/mol

Convert unit into Kg/mol

28 g/mol x 1 Kg/1000g = 2.8x10¯² Kg/mol

Convert temperature into kelvin

T = 25 + 273 = 298K

Put all values in formula

Vrms = SQRT (3RT/M)

Vrms = SQRT (3x8.314x298/2.8x10¯²)

Vrms = SQRT (265454. 14)

Vrms = 515.22 m/s

5 0
3 years ago
A skydiver jumps out of a plane and immediately begins falling toward the Earth.
worty [1.4K]

Answer:

5.2m/s^2

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)\\{Vf^{2}-Vo^2}/{2.a} =X(2)\\X= VoT+0.5at^{2} (3)\\X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this case we can use the ecuation number 3

x=100m

t=6.2s

Vo=0m/s

X= VoT+0.5at^{2}\\X= 0.5at^{2}\\a=\frac{X}{0.5t^2} \\a=\frac{100}{0.5(6.2)^2}=5.2m/s^2

7 0
3 years ago
A 1500 kg sedan goes through a wide intersection traveling from north to south when it is hit by a 2500 kg SUV traveling from ea
kolezko [41]

Answer:

v = 19.33 m / s   South

Explanation:

To solve this exercise we must use the conservation of momentum, for which we must define a system formed by the two cars, therefore the forces during the collision are internal and therefore the moment is conserved.

Since it is a vector quantity, we are going to work on each axis, the x axis is in the East-West direction

initial instant. Before the crash

        p₀ = m 0 + M v₂ₓ

final instant. Right after the crash

        p_f = (m + M) vₓ

        p₀ = 0_pf

        M v₂ₓ = (m + M) vₓ

In this case m is the mass of the car and M the mass of the SUV

        vₓ = \frac{M}{m+My}  v₂ₓ            (1)

in the Y axis (North - South direction)

initial instant

       p₀ = m v_{1y} + M 0

final moment

       p_f = (m + M) v_y

       p₀ = p_f

       m v_{1y} + M 0 = (m + M) v_y

       v_y = \frac{m}{m+M}  \  v_{1y}       (2)

With these speeds we can use the relationship between work and the variation of kinetic energy, in this part the two cars are already united.

         W = ΔK

friction force work is

         W = - fr d

the friction force is described by the equation

         fr = μ N

Newton's second law

         N-W = 0

         N = W

         

we substitute

         W = - μ (m + M) g d

as the car stops the final kinetic energy is zero and

the initial kinetic energy is

         K₀ = ½ (m + M) v²

we substitute

         - μ (m + M) g d = 0 - ½ (m + M) v²

            μ g d = ½ v²

            v² = 2 μ g d

the distance traveled can be found with the Pythagorean theorem

        d = \sqrt{x^2+y^2}

        d = \sqrt{5.48^2 + 6.37^2}

        d = 8.40 m

let's calculate the speed

         v² = 2 0.75 9.8 8.40

         v = √123.48

         v = 11.11 m / s

this velocity is in the direction of motion so we can use trigonometry to find the angles

          tan θ = y / x

          θ = tan⁻¹ y / x

          θ = tan⁻¹ (-5.48 / -6.37)

          θ = 40.7º

Since the two magnitudes are negative, this angle is in the third quadrant, measured from the positive side of the x-axis in a counterclockwise direction.

          θ'= 180 + 40.7

          θ’= 220.7º

In the exercise they indicate the the sedan moves in the y-axis, therefore

          sin θ'= v_y / v

          v_y = v sin 220.7

          v_y = 11.11 sin 220.7

          v_y = -7.25 m / s

the negative sign indicates that it is moving south

To find the speed we substitute in equation 2

          v_y = \frac{m}{m+M}  \ v_{1y}

          v_{1y} = v_ y   \frac{m+M}{m}

           

let's calculate

         v_{1y} = -7.25    \frac{1500+2500}{1500}

         v_{1y} = - 19.33 m/s

therefore the speed of the sedan is v = 19.33 m / s with a direction towards the South

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3 years ago
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6 0
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graphs are used to show how the responding variable changes in response to the -------------------  variable.
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The answer is the manipulative variable.  y=kx (equation for a straight line) The responding variable, y, changes because of the manipulated variable, x.  k is the slope
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