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Mashcka [7]
3 years ago
11

A scuba diver wears weights as well as a buoyancy compensator to establish neutral buoyancy while diving. The buoyancy compensat

or can either be inflated with air or the air in it can be released. Explain how a scuba diver uses the buoyancy compensator to dive and to rise back to the surface.
Physics
1 answer:
katrin [286]3 years ago
4 0

Answer:

With more air is more buoyancy. When deflated or released the scuba diver is less buoyant.

Explanation:

The compensator is a Buoyancy control device that has an inflatable air bladder.When we have more air out into the inflatable bladder, then one is more buoyant. If the air is released from the bladder, then one is less buoyant. We add air through an air inflation valve. Air is also then released using air-deflation valves.

Buoyancy can be defined as an upward force which is exerted on an object that is fully or partially immersed in water

when one is less buoyant than water, it means that the upward pressure is more than the downward pressure of that person and his equipment. Then he will float. In a case of negative buoyancy, we have downward pressure of this person and his equipment to be more than the upward pressure of the water. Then sinking will happen.

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weqwewe [10]

Answer:

75j

Explanation:

4 0
3 years ago
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A wire 3.22 m long and 7.32 mm in diameter has a resistance of 11.9 mΩ. A potential difference of 33.7 V is applied between the
Scorpion4ik [409]

Answer:

(a) Current is 2831.93 A

(b) 8.40A/m^2

(c) \rho =15.52\times 10^{-9}ohm-m

Explanation:

Length of wire l = 3.22 m

Diameter of wire d = 7.32 mm = 0.00732 m

Cross sectional area of wire

A=\pi r^2=3.14\times 0.00366^2=4.20\times 10^{-5}m^2

Resistance R=11.9mohm=11.9\times 10^{-3}ohm

Potential difference V = 33.7 volt

(A) current is equal to

i=\frac{V}{R}=\frac{33.7}{11.9\times 10^{-3}}=2831.93A

(B) Current density is equal to

J=\frac{i}{A}

J=\frac{2831.93}{4.20\times 10^{-5}}=8.40A/m^2

(c) Resistance is equal to

R=\frac{\rho l}{A}

11.9\times 10^{-3}=\frac{\rho \times 3.22}{4.20\times 10^{-5}}

\rho =15.52\times 10^{-9}ohm-m

4 0
3 years ago
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 93.3 V/m. Calculate the intensity S of this wave
7nadin3 [17]

Answer:

Intensity = 11.56W/m²

The energy flowing through the given area is 4.55 J

Explanation:

The expression for the intensity of the electromagnetic wave is,

I = \frac{1}{2} C{ {\varepsilon _0}E_m^2

Here,\varepsilon _0 is the permittivity of the free space,

E_m  is the electric field amplitude and

c is the speed of the light.

substitute

⁸m/s for c

8.85×10  −12  C² /N⋅m² for {\varepsilon _0}

and 93.3 V/m for {E_{\rm{m}

I = \frac{1}{2} \times (3\times10^8)\times(8.85\times10^-^1^2)(93.3)\\\\I = 11.56W/m^2

The expression for the energy is,

E = I×A×t

Here, I is the intensity of the electromagnetic wave,

A is the area, and

t is the time.

Substitute

11.56W/m² for I

0.0287m ² for A

13.7s for t

E = (11.56)\times(0,0287)\times(13.7)\\E = 4.55J

The energy flowing through the given area is 4.55 J

5 0
4 years ago
What unit is electrical energy typically measured in on energy bills?
velikii [3]
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3 years ago
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What is the formula for calculating acceleration​
Komok [63]

Answer:

the formula for calculating acceleration is ending speed minus starting speed divided by time.

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