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Montano1993 [528]
3 years ago
11

Scuba divers are warned that if they must make a rapid ascent, they should exhale on the way up. If a diver rapidly ascends to t

he surface with lungs full of air, his lungs could be damaged. Explain why this is so.
Physics
1 answer:
Whitepunk [10]3 years ago
5 0

Air for a diver comes out of a high pressure tank at - Same- pressure compared to the water around the diver (metered by the regulator).

This means the lungs are inflated with - Highly pressurized- gas.

This does not adversely affect the diver when deep underwater, because the entire environment around the diver is at -Same - pressure.

If the diver suddenly surface, the air in the alveoli in the lungs will still be at - a higher - pressure compared to the air around the diver, which will be at - a lower - pressure.

The gas in the diver's lungs will - expand - and can damage the alveoli.

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This is known as work
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In an experiment, you find the mass of a cart to be 250 grams. What is the mass of the cart in kilograms?
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Read 2 more answers
(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy r
Nikitich [7]

Answer:

Explanation:

Gravitational Potential Energy at earth surface U_1=\frac{GM_em}{R_e}

Gravitational Potential Energy at height h is U_2=\frac{GM_em}{R_e+h}

Energy required to lift the satellite E_1=U_1-U_2

E_1=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}

Now Energy required to orbit around the earth

E_2=\frac{1}{2}mv_{orbit}^2=\frac{GM_2m}{2(R_e+h)}

\Delta E=E_1-E_2

\Delta E=\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}

E_1=E_2  (given)

\frac{GM_em}{R_e}-\frac{GM_em}{R_e+h}-\frac{GM_2m}{2(R_e+h)}=0

\frac{1}{R_e}-\frac{3}{2(R_e+h)}=0

h=\frac{R_e}{2}

h=3.19\times 10^6\ m

(b)For greater height E_1  is greater than E_2

thus energy to lift the satellite is more than orbiting around earth

4 0
3 years ago
Two lasers, one red (with wavelength 633.0 nmnm) and the other green (with wavelength 532.0 nmnm), are mounted behind a 0.150-mm
Ratling [72]

(a) The screen  is 3.20m from the split.

(b) The closest minima for green, distance Δy = 0.45 cm.

When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.

(a)Equation of minima = sinθ  = mλ/α

Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015

Putting the values in formula to get θ.

  θ = sin⁻¹ ( \frac{3 X 6.33X10^{-7} }{0.00015} ) = 0.01266 rad

triangle need to be drawn to find relationship between θ, y$ and L

tan(θ) = y/L  where; y = 4.05 cm

L = y/tan(θ) = 3.20

Hence, the screen is 3.20m from the split.

(b) Find the closest minima for green

minima equation is sinθ  = mλ/α where, m = 4 (minima with smallest distance)

sinθ  = 4λ/α

θ = sin⁻¹ (\frac{4X6.33X10^{-7} }{0.00015}) = 0.01688 rad

Calculate L using

tanθ = y/L

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From equation subtract y₃ from y:

                 4.50 cm - 4.05 cm = 0.45 cm

Hence, distance Δy = 0.45 cm.

Learn more about the Diffraction with the help of the given link:

brainly.com/question/12290582

#SPJ4

I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot

Question

Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.

a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.

b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?

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1 year ago
If a lamp has a resistance of 50 ohms and is operated by a p.d. of 10V, find the current flowing through it
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V=IR therefore I=V/R=10/50=0.2A therefore the current is 0.2 A
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