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raketka [301]
2 years ago
12

In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t

hat the maximum pressure difference his lungs can manage and still breathe is -73 mm-Hg. 1 mm-Hg = 133 N/m2.
Determine the maximum distance below the water surface, at which Tarzan could breathe with the reed.
Physics
1 answer:
gladu [14]2 years ago
8 0

Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

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A spaceship is headed toward Alpha Centauri at 0.999c. According to us, the distance to Alpha Centauri is about 4 light-years. H
aniked [119]

Answer:

According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>

<em></em>

Explanation:

For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:

v = d / t

where

  • v is the speed of the spaceship
  • d is the distance
  • t is the time

Therefore,

d = v × t

d = (0.999 c)(4 light-years)

d = 3.996  light-years

This distance is<em> very slightly less than 4 light-years. </em>

4 0
2 years ago
If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water
Svetradugi [14.3K]

Answer:

The air-water interface is an example of<em> </em>boundary. The <u><em>transmitted</em></u><em> </em> portion of the initial wave energy is way smaller than the <u><em>reflected</em></u><em> </em> portion. This makes the <u><em>boundary</em></u>  wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can <u><em>travel directly to your ear</em></u>.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

4 0
2 years ago
An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr
AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity  (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

a = \dfrac{400}{2\times 50}

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

\mu = \dfrac{a}{g}

\mu = \dfrac{4}{9.81}

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0
2 years ago
A ball on a string travels once around a circle with a circumference of 2.0 m. The tension in the string is 5.0 N. how much work
SpyIntel [72]

Answer:0

Explanation:

Given

circumference of circle is 2 m

Tension in the string T=5 N

2\pi r=2

r=\frac{2}{2\pi }=\frac{1}{\pi }=0.318 m

In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is 90^{\circ}

W=\int\vec{F}\cdot \vec{r}

W=\int Fdr\cos 90

W=0

4 0
1 year ago
The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune
Veronika [31]
Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
\frac{r^3}{T^2}= \frac{GM}{4 \pi^2} (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
1 AU = 1.5 \cdot 10^{11} m
so the radius of the orbit is
r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
T=\sqrt{ \frac{4 \pi^2}{GM} r^3} =  \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years
3 0
1 year ago
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