1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Akimi4 [234]
3 years ago
13

Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snork

el tube and thus are at atmospheric pressure. In atmospheres, what is the difference Δp between this internal air pressure and the water pressure against the body if the length of the snorkel tube is
(a)24 cm (standard situation) and



(b)4.1 m (probably lethal situation)?
Physics
1 answer:
fgiga [73]3 years ago
7 0

Answer:

2354.4 Pa

40221 Pa

Explanation:

\rho = Density = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth

The pressure difference would be

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa

The pressure difference in the first case is 2354.4 Pa

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa

The pressure difference in the second case is 40221 Pa

You might be interested in
The Falcon 9 rocket has a mass of 549,000 kg and in 70 seconds into the launch, the rocket reaches a speed of 343.2 m/s. What is
yaroslaw [1]

Explanation:

=5.49*10^5*343.2

=1.8846*10^8

8 0
2 years ago
Three equal point charges, each with charge 1.45 μCμC , are placed at the vertices of an equilateral triangle whose sides are of
LUCKY_DIMON [66]

Answer:

U = 80.91 J

Explanation:

In order to calculate the electric potential energy between the three charges you use the following formula:

U=k\frac{q_1q_2}{r_{1,2}}                  (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q1: q2 charge

r1,2: distance between charges 1 and 2.

For the three charges you have:

U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}           (2)

You use the fact that q1=q2=q3=q and that the distance between charges are equal. Then, in the equation (2) you have:

q = 1.45μC = 1.45*10^-6C

r = 0.700mm = 0.700*10^-3m

U_T=3k\frac{q^2}{r}=3(8.98*10^9Nm^2/C^2)\frac{(1.45*10^{-6}C)}{0.700*10^{-3}m}\\\\U_T=80.91J

The electric potential energy between the three charges is 80.91 J

7 0
3 years ago
A rectangular sharp-crested weir is contracted on both sides, and the opening is 1.2 m wide. At what height (Hw) should it be pl
Alex

Answer:

H_w = 2.129 m

Explanation:

given,

Width of the weir, B = 1.2 m

Depth of the upstream weir, y = 2.5 m

Discharge, Q = 0.5 m³/s

Weir coefficient, C_w = 1.84 m

Now, calculating the water head over the weir

Q = C_w BH^{3/2}

H = (\dfrac{Q}{C_wB})^{2/3}

H = (\dfrac{0.5}{1.84\times 1.2})^{2/3}

H = 0.371\ m

now, level of weir on the channel

H_w = y - H

H_w = 2.5 - 0.371

H_w = 2.129 m

Height at which weir should place is equal to 2.129 m.

7 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
Why are there only two elements in the first period of the periodic table?(1 point)
tester [92]

Answer:

because each row increases in atomic mass by a specific number, so anything over five is in the second row.

8 0
2 years ago
Other questions:
  • which organelle in the plant cell would mainly help the cell tack in water or get rid of water just like the potato did? this is
    14·1 answer
  • What are nasa's four great observatories and in what parts of the electromagnetic spectrum do they observe?
    5·2 answers
  • If the tensile strength of the Kevlar 49 fibers is 0.550 x 106 psi and that of epoxy resin is 11.0 x103 psi, calculate the stren
    5·1 answer
  • A sample neon gas has its volume tripled and its temperature held constant. What will be the new pressure relative to the initia
    5·1 answer
  • what’s 55mph to km/min? can someone explain to to me with the work so i can understand how to solve this
    12·2 answers
  • Force causes __________ dimensional objects to rotate
    6·1 answer
  • Which is an example of chemical energy to thermal energy?
    13·2 answers
  • Please help!!! This Is for science
    11·1 answer
  • WILL MARK BRAINLIEST!!!!!<br> How is the 3rd law different from the 1st and 2nd laws?
    15·1 answer
  • Explain how conduction, convection, and radiation occur involving a campfire
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!