Question

1. Engineers have developed a bathyscaph that can reach ocean depths of 7.0 mi. [1 mile = 1.6 km] (a) What is the absolute pressure at this depth? (b) If the inside of this vessel is maintained at atmospheric pressure, what is the net force on a porthole of diameter 15 cm? 2. Visualize a collapsible bag filled with a glucose solution that is connected to an IV. The bag is hung on a stand over a hospital bed (the IV obviously is inserted into the patient's arm). If the average gauge pressure in the patient's artery is 1.33 x 10^4 Pa, what must be the minimum height of the hanging bag in order to force the glucose solution into the patient's arm? Assume the average density of the glucose solution to be 1020 kg/m^3. 7. A hydraulic automobile lift consists of an input piston with an area of 64.0 cm^2 and an output piston face of 3.20 x 10^3 cm^2. If a 900. kg car rests on the latter, a. how much force must be exerted to raise the vehicle very slowly. weight ? w = mg = m(9.80 m/s^2) ? kg m/s^2 = Newton b. what is the internal fluid pressure? A_in = 64.0 cm^2 A_out = 3.20 x 10^3 cm^2 (not drawn to scale)

          1. Engineers have developed a bathyscaph that can reach ocean depths of 7.0 mi. [1 mile =
1.6 km]

(a) What is the absolute pressure at this depth?
(b) If the inside of this vessel is maintained at atmospheric pressure, what is the net
force on a porthole of diameter 15 cm?

2. Visualize a collapsible bag filled with a glucose solution that is connected to an IV.
The bag is hung on a stand over a hospital bed (the IV obviously is inserted into the
patient's arm). If the average gauge pressure in the patient's artery is 1.33 x 10^4 Pa,
what must be the minimum height of the hanging bag in order to force the glucose
solution into the patient's arm? Assume the average density of the glucose solution
to be 1020 kg/m^3.

7. A hydraulic automobile lift consists of an input piston with an area of 64.0 cm^2 and an
output piston face of 3.20 x 10^3 cm^2. If a 900. kg car rests on the latter,
a. how much force must be exerted to raise the vehicle very slowly.
weight ? w = mg = m(9.80 m/s^2) ? kg m/s^2 = Newton
b. what is the internal fluid pressure?

A_in = 64.0 cm^2
A_out = 3.20 x 10^3 cm^2
(not drawn to scale)

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1. Engineers have developed a bathyscaph that can reach ocean depths of 7.0 mi. [1 mile =
1.6 km]

(a) What is the absolute pressure at this depth?
(b) If the inside of this vessel is maintained at atmospheric pressure, what is the net
force on a porthole of diameter 15 cm?

2. Visualize a collapsible bag filled with a glucose solution that is connected to an IV.
The bag is hung on a stand over a hospital bed (the IV obviously is inserted into the
patient's arm). If the average gauge pressure in the patient's artery is 1.33 x 10^4 Pa,
what must be the minimum height of the hanging bag in order to force the glucose
solution into the patient's arm? Assume the average density of the glucose solution
to be 1020 kg/m^3.

7. A hydraulic automobile lift consists of an input piston with an area of 64.0 cm^2 and an
output piston face of 3.20 x 10^3 cm^2. If a 900. kg car rests on the latter,
a. how much force must be exerted to raise the vehicle very slowly.
weight ? w = mg = m(9.80 m/s^2) ? kg m/s^2 = Newton
b. what is the internal fluid pressure?

Ain = 64.0 cm^2
Aout = 3.20 x 10^3 cm^2
(not drawn to scale)

Added by Alyssa H.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Cyra Jelle Calleja

12/07/2021

Step 1

0 miles in the ocean. Given: - Atmospheric pressure = 1.013 x 10^5 Pa - Density of seawater = 1024 kg/m^3 - Acceleration due to gravity = 9.8 m/s^2 - Depth = 7.0 miles = 7.0 x 1.6 km = 11.2 km = 11,200 m Absolute pressure = Atmospheric pressure + (density of fluid Show more…

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1. Engineers have developed a bathyscaph that can reach ocean depths of 7.0 mi. [1 mile = 1.6 km] (a) What is the absolute pressure at this depth? (b) If the inside of this vessel is maintained at atmospheric pressure, what is the net force on a porthole of diameter 15 cm? 2. Visualize a collapsible bag filled with a glucose solution that is connected to an IV. The bag is hung on a stand over a hospital bed (the IV obviously is inserted into the patient's arm). If the average gauge pressure in the patient's artery is 1.33 x 10^4 Pa, what must be the minimum height of the hanging bag in order to force the glucose solution into the patient's arm? Assume the average density of the glucose solution to be 1020 kg/m^3. 7. A hydraulic automobile lift consists of an input piston with an area of 64.0 cm^2 and an output piston face of 3.20 x 10^3 cm^2. If a 900. kg car rests on the latter, a. how much force must be exerted to raise the vehicle very slowly. weight → w = mg = m(9.80 m/s^2) → kg m/s^2 = Newton b. what is the internal fluid pressure? A_in = 64.0 cm^2 A_out = 3.20 x 10^3 cm^2 (not drawn to scale)

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00:01 Okay, so question 1a asks for the absolute pressure.

00:04 The formula for absolute pressure is this.

00:08 Atmospheric pressure plus the product of the density of the fluid, the acceleration due to gravity in the height or depth.

00:17 So that atmospheric pressure is 1 .013 times 10 raised to 5 pascal.

00:23 So the density of the fluid, in this case the fluid is ocean, so the seawater has a density of 1 ,024 kilogram per meter cube.

00:35 The acceleration due to gravity is 9 .8 meter per second squared, and we are giving the depth of 7 miles.

00:41 So that is 7 times 1 .6 kilometer.

00:45 So 1 .6 kilometer is 1 ,600 meters.

00:50 Solving, we can find that the absolute pressure is 1 .125 times 10 raised to 8 pascal.

01:03 Now, question be asked for the neb force on the porthole that is a diameter of 15 cm if the inside of the vessel is maintained at atmospheric pressure.

01:16 So since the formula for pressure is force over area, then force is equals to the product of this force, of this pressure, and the area of the porthole, where this one is just the difference of the absolute pressure in the atmospheric pressure.

01:43 So, an answer to a minus the atmospheric pressure.

01:47 Then we have, so the radius is half of the diameter.

01:59 We are given the diameter.

02:07 So we have this.

02:08 Then solving, you have 1 .125 times 10 raised to 8 pascal's minus 1 .013 times 10 raised to 5 pascal.

02:23 Note that this pressure is called gauge pressure.

02:26 So this one is gauge pressure.

02:28 This one is gauge pressure.

02:32 So 15 centimeters in meters is 0 .15.

02:43 In solving, we can find that the force is 1 .986 times 10 raised to 6 tons.

02:52 Now question 2 as for the height.

02:59 You can determine that from the formula for gauge pressure.

03:05 Which is this, and isolating the height to the left side.

03:11 We have this...

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