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Dmitrij [34]
3 years ago
6

What is the external-internal pressure difference when the diver's lungs are at a depth of 6.1m (about 20ft)? assume that the di

ver is in fresh water. (a scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)?
Physics
1 answer:
Nesterboy [21]3 years ago
8 0
Snorkel connects the diver's lungs with the air above the water. Because of this, the pressure inside the diver's lungs is the same as the atmospheric pressure.
The static pressure of a fluid is given with this equation:
P=\rho gh
The density of water is 1000kg/m^3.
P=1000\cdot 9.81 \cdot 6.1=59841$Pa
Standart atmospheric pressure is 101<span>325Pa. 
</span>The difference is:
\Delta P=101325-59841=41484 $Pa

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A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br
Levart [38]

Answer:

(A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

\lambda=\dfrac{dy}{mD}

m=\dfrac{d y}{\lambda D}

Put the value into the formula

m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}

m=5.65 = 6

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

\beta=\dfrac{yd}{D}

Put the value in to the formula

\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}

\beta=3.33\times10^{-6}\ m

\beta=3.33\ \mu m

Hence, (A).  The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0
2 years ago
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butalik [34]

Answer:

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ω=θ/t and it SI unit is rad/s

Explanation:

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Answer:

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Is NaCH an element? Why or why not?
amm1812
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