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1 year 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.

1 answer:

gladu [14]1 year 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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Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

6 0

1 year ago

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

pickupchik [31]

The optimal angle of 45° for maximum horizontal range is only valid when initial height is the same as final height.

<span>In that particular situation, you can prove it like this: </span>

<span>initial velocity is Vo </span>
<span>launch angle is α </span>

<span>initial vertical velocity is </span>
<span>Vv = Vo×sin(α) </span>

<span>horizontal velocity is </span>
<span>Vh = Vo×cos(α) </span>

<span>total time in the air is the the time it needs to fall back to a height of 0 m, so </span>
<span>d = v×t + a×t²/2 </span>
<span>where </span>
<span>d = distance = 0 m </span>
<span>v = initial vertical velocity = Vv = Vo×sin(α) </span>
<span>t = time = ? </span>
<span>a = acceleration by gravity = g (= -9.8 m/s²) </span>
<span>so </span>
<span>0 = Vo×sin(α)×t + g×t²/2 </span>
<span>0 = (Vo×sin(α) + g×t/2)×t </span>
<span>t = 0 (obviously, the projectile is at height 0 m at time = 0s) </span>
<span>or </span>
<span>Vo×sin(α) + g×t/2 = 0 </span>
<span>t = -2×Vo×sin(α)/g </span>

<span>Now look at the horizontal range. </span>
<span>r = v × t </span>
<span>where </span>
<span>r = horizontal range = ? </span>
<span>v = horizontal velocity = Vh = Vo×cos(α) </span>
<span>t = time = -2×Vo×sin(α)/g </span>
<span>so </span>
<span>r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) </span>
<span>r = -(Vo)²×sin(2α)/g </span>

<span>To find the extreme values of r (minimum or maximum) with variable α, you must find the first derivative of r with respect to α, and set it equal to 0. </span>

<span>dr/dα = d[-(Vo)²×sin(2α)/g] / dα </span>
<span>dr/dα = -(Vo)²/g × d[sin(2α)] / dα </span>
<span>dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα </span>
<span>dr/dα = -2 × (Vo)² × cos(2α) / g </span>

<span>Vo and g are constants ≠ 0, so the only way for dr/dα to become 0 is when </span>
<span>cos(2α) = 0 </span>
<span>2α = 90° </span>
<span>α = 45° </span>

4 0

1 year ago

A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir

ikadub [295]

In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

KE1 = KE2

The kinetic energy of the system before the collision is solved below.

KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s.

6 0

1 year ago

A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte

Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

= 3/2×(5.3)×(8.31451)×(24 + 273)

= 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0

1 year ago

A turntable rotates counterclockwise at 76 rpm . A speck of dust on the turntable is at 0.47 rad at t=0. What is the angle of th

icang [17]

To solve this exercise it is necessary to apply the kinematic equations of angular motion.

By definition we know that the displacement when there is constant angular velocity is

$\theta= \theta_0 +\omega t$

From our given data we know that,

$\omega = 76\frac{rev}{min}$

$\omega = 76\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1 min}{60s})$

$\omega = 7.958rad/s$

Moreover we know that

$\theta_0 = 0.47 rad$

Therefore for time t=8.1s we have,

$\theta= \theta_0+ \omega t$

$\theta= 0.47+(7.958)(8.1)$

$\theta = 64.9298rad$

That number in revolution is:

$\theta = 64.9298rad(\frac{1rev}{2\pi})$

$\theta = 15.108 Revolutions$

Here, we see that there are 15 complete revolutions

And 0.108 revolutions i not complete, so the tunable rotation is

$\theta_{net} = 0.108*2\pi=0.216\pi$

Therefore the angle of the speck at a time 8.1s is $0.216\pi$

4 0

2 years ago