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elena55 [62]
2 years ago
5

If a scuba diver fills his lungs to full capacity of 5.5 L when 10 m below the surface, to what volume would his lungs expand if

he quickly rose to the surface? Assume, the temperature of the air inside the lungs remains costant. Density of sea water is 1025 kg/m3.
Physics
2 answers:
kakasveta [241]2 years ago
8 0

Answer:

10.95 L

Explanation:

Let the pressure at depth 10 m is P1.

V1 = 5.5 L

P1 = Po + h x d x g

P2 is the pressure at the surface

P2 = Po

where, Po be the atmospheric pressure  = 1.01 x 10^5 Pa

Let V2 be the volume at the surface.

Use P1 x V1 = P2 x V2

(Po + 10 x 1025 x 9.8) x 5.5 = Po x V2

(1.01 x 10^5 + 1 x 10^5) x 5.5 = 1.01 x 10^5 x V2

2.01 x 5.5 = 1.01 x V2

V2 = 10.95 L

umka21 [38]2 years ago
3 0

Answer:

The volume at the surface is 10.97 L.

Explanation:

Given that,

Volume = 5.5 L

Height = 10 m

Density of sea water= 1025 kg/m³

We need to calculate the pressure at that point

Using formula of pressure

P'=P+\rho gh

Put the value into the formula

P'=1.01\times10^{5}+1025\times9.8\times10

P'=201450\ Pa

We need to calculate the volume at the surface

Using equation of ideal gas

PV= RT

So, for both condition

PV=P'V'

Put the value into the formula

V=\dfrac{201450\times5.5}{1.01\times10^{5}}

V=10.97\ L

Hence, The volume at the surface is 10.97 L.

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A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.
Katen [24]

Answer:

9.4 m/s

Explanation:

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy of the object.

So we can write:

W=K_f - K_i

where in this problem:

W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)

K_i = 66,120 J is the initial kinetic energy of the car

K_f is the final kinetic energy

Solving for Kf,

K_f = W+K_i = -36,733+66,120=29,387 J

The kinetic energy of the car can be also written as

K_f = \frac{1}{2}mv^2

where:

m = 661 kg is the mass of the car

v is its final speed

Solving, we find

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s

8 0
3 years ago
At which of the above boundries is sea floor destroyed
Slav-nsk [51]
The continental crust
Hope it helps!!
7 0
3 years ago
Which nucleus completes the following equation?<br> A. 299 Np<br> B. 20Pa<br> C. 2 Pa<br> D. - Np
blondinia [14]

Answer:

Option D. ²³⁹₉₃Np

Explanation:

Let the unknown be ʸₓA.

Thus, the equation becomes:

²³⁹₉₂U —> ⁰₋₁e + ʸₓA

Next, we shall determine the x, y and A. This can be obtained as follow:

92 = –1 + x

Collect like terms

92 + 1 = x

93 = x

x = 93

239 = 0 + y

239 = y

y = 239

ʸₓA => ²³⁹₉₃A => ²³⁹₉₃Np

Thus, the complete equation is:

²³⁹₉₂U —> ⁰₋₁e + ²³⁹₉₃Np

7 0
3 years ago
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Katyanochek1 [597]

1. 2 way radio

2. radio waves

3. communication

4. convert the voltage from a transmitter into a radio signal

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4 0
3 years ago
A 97.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.63 r
sammy [17]

Answer:

the final angular velocity of the platform with its load is 1.0356 rad/s

Explanation:

Given that;

mass of circular platform m = 97.1 kg

Initial angular velocity of platform ω₀ = 1.63 rad/s

mass of banana m_{b} = 8.97 kg

at distance r = 4/5  { radius of platform }

mass of monkey m_{m} = 22.1 kg

at edge = R

R = 1.73 m

now since there is No external Torque

Angular momentum will be conserved, so;

mR²/2 × ω₀ = [ mR²/2 + m_{b} (\frac{4}{5} R)² + m_{m}R² ]w

m/2 × ω₀ = [ m/2 + m_{b} (\frac{4}{5} )² + m_{m} ]w

we substitute

w = 97.1/2 × 1.63 / ( 97.1/2 + 8.97(16/25) + 22.1

w = 48.55 × [ 1.63 / ( 48.55 + 5.7408 + 22.1 )

w = 48.55 × [ 1.63 / ( 76.3908 ) ]

w = 48.55 × 0.02133

w = 1.0356 rad/s

Therefore; the final angular velocity of the platform with its load is 1.0356 rad/s

8 0
3 years ago
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