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beks73
8 days ago
12

If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil

l it last at 20 metres/66 feet breathing normally? 60 minutes
Physics
1 answer:
kicyunya [3.1K]8 days ago
8 0

Response:

Clarification:

Consider the air volume as V, at standard atmospheric pressure, which is 10⁵ Pa

At a depth of 20 meters, the pressure is computed as

standard pressure + hdg

10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa

Under this pressure, the volume V is reduced to V/ 2.96

This reduced volume translates to lasting for 1/2.96 the original time, or 60/2.96 = 20.27 minutes.

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The amount of kinetic energy an object has depends on its mass and its speed. Rank the following sets of oranges and cantaloupes
Sav [3045]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>
6 0
1 month ago
An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t
Maru [3263]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m

The worker is positioned at 2.5 m

7 0
1 month ago
A rocket starts from rest and moves upward from the surface of the earth. for the first 10.0 s of its motion, the vertical accel
kicyunya [3158]

The rocket's acceleration is described here as

a_y = 2.60* t

now recognizing that

\frac{dv}{dt} = 2.60t

we integrate both sides

\int dv = \int 2.60t dt

v = 2.60\frac{t^2}{2}

v = 1.30 t^2

given that the rocket is accelerating for a duration of t = 10 s

thus, we have

v = 1.30 * 10^2

v = 130 m/s

consequently, after t = 10 s, the rocket will achieve a speed of 130 m/s in an upward direction

5 0
13 days ago
Sully is riding a snowmobile on a flat, snow-covered surface with a constant velocity of 10 meters/second. The total mass of the
inna [2995]
Result: 168N

The calculation shows 16 - 10 equals 6
and 6 divided by 10 equals 0.6
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7 0
13 days ago
Read 2 more answers
A man makes a 27.0 km trip in 16 minutes. (a.) How far was the trip in miles? (b.) If the speed limit was 55 miles per hour, wa
Maru [3263]

Answer:

(a) 16.777 miles

(b) Yes, he exceeded the speed limit

Explanation:

(a)

We need to perform the necessary calculations to convert kilometers to miles:

27km*\frac{1000m}{1km} *\frac{1mi}{1609.34m} =16.77706389mi

Thus, the distance of the trip in miles is:

d=16.77706389mi

(b)

Next, we will compute the man's speed during the journey:

v=\frac{d}{t}

Before that, we must convert minutes to hours:

16min*\frac{1h}{60min} =2.666666667h

The resulting speed is:

v=\frac{16.77706389mi}{2.666666667h} =62.91398959\frac{mi}{h}

Consequently:

62.91398959\frac{mi}{h}>55\frac{mi}{h}

Thus, it can be concluded that the driver was speeding

8 0
1 month ago
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