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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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What evidence contradicts the theory that moving two-ton stones was physically impossible for the egyptians, and must have been
SOVA2 [1]
Archaeological evidence shows that Egyptians worked together to build the pyramids. Remains of quarries and ancient tools suggest that large slabs were created from rock beds. The slabs were placed on sleds and pulled to the building site. To make this process easier, men most likely poured oil on the roadway. This process is depicted in tomb paintings that date back to 1900 BCE.
8 0
2 years ago
Una cuerda de violin vibra con una frecuencia fundamental de 435 Hz. Cual sera su frecuencia de vibracion si se le somete a una
EleoNora [17]

Answer:

a)  f = 615.2 Hz      b)  f = 307.6 Hz

Explanation:

The speed in a wave on a string is

         v = √ T / μ

also the speed a wave must meet the relationship

          v = λ f

           

Let's use these expressions in our problem, for the initial conditions

            v = √ T₀ /μ

             √ (T₀/ μ) = λ₀ f₀

now it indicates that the tension is doubled

         T = 2T₀

          √ (T /μ) = λ f

          √( 2To /μ) = λ f

         √2  √ T₀ /μ = λ f

we substitute

         √2 (λ₀ f₀) = λ f

if we suppose that in both cases the string is in the same fundamental harmonic, this means that the wavelength only depends on the length of the string, which does not change

           λ₀ = λ

           f = f₀ √2

           f = 435 √ 2

           f = 615.2 Hz

b) The tension is cut in half

         T = T₀ / 2

         √ (T₀ / 2muy) =  f = λ f

          √ (T₀ / μ)  1 /√2 = λ f

           fo / √2 = f

           f = 435 / √2

           f = 307.6 Hz

Traslate

La velocidad en una onda en una cuerda es

         v = √ T/μ

ademas la velocidad una onda debe cumplir la relación

          v= λ f  

           

Usemos estas expresión en nuestro problema, para las condiciones iniciales

            v= √ To/μ

             √ ( T₀/μ) = λ₀ f₀

ahora nos indica que la tensión se duplica

         T = 2T₀

          √ ( T/μ) = λf

          √ ) 2T₀/μ = λ f

         √ 2 √ T₀/μ = λ f

         

substituimos  

         √2    ( λ₀ f₀)  =  λ f

si suponemos que en los dos caso la cuerda este en el mismo armónico fundamental, esto es que la longitud de onda unicamente depende de la longitud de la cuerda, la cual no cambia

                 λ₀ =  λ

           f = f₀ √2

           f = 435 √2

           f = 615,2 Hz

b)  La tension se reduce a la mitad

         T = T₀/2    

         RA ( T₀/2μ)  =  λ  f

          Ra(T₀/μ) 1/ra 2  =  λ f

           fo /√ 2 = f

           f = 435/√2

           f = 307,6 Hz

5 0
1 year ago
Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0
2 years ago
While on a camping trip, a man would like to carry water from the lake to his campsite. He fills two, non‑identical buckets with
xenn [34]

Answer:

The bucket  located in the direction opposite of displacement of the rod for balancing will be heavier.

Explanation:

Moment of a force is directly proportional to it's distance from the point about which rotation takes place. In the given case the man increases the lever arm(distance from point of rotation of the force) to balance the moment of the lighter bucket. Thus in the direction of the increase the lighter bucket will be placed.

6 0
2 years ago
Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of
Aleks [24]

We can solve the problem by using Kepler's third law, which states:

\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}

where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.

Using the data of the problem:

T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s

r=384000 km=3.84 \cdot 10^8 m

We can re-arrange the equation and find the Earth's mass:

M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg

4 0
2 years ago
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