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4 years ago

Predict using Boyle's law, what will happen to a balloon that an ocean diver takes to a pressure of 202 kPa.

Physics

1 answer:

kobusy [5.1K]4 years ago

6 0

The volume of the balloon will halve

Explanation:

Boyle's law states that for an ideal gas kept at constant temperature, the pressure of the gas is proportional to its volume. Mathematically,

$pV=const.$

where

p is the gas pressure

V is the volume

The equation can also be rewritten as

$p_1 V_1 = p_2 V_2$

And if we apply it to the gas inside the balloon in this problem (assuming its temperature is constant), we have:

$p_1 = 101 kPa$ is the initial pressure at sea level (the atmospheric pressure)

$V_1$ is the initial volume

$p_2 = 202 kPa$ is the final pressure

$V_2$ is the final volume

Substituting into the equation, we find:

$V_2 = \frac{p_1 V_1}{p_2}=\frac{(101)V_1}{202}=\frac{V_1}{2}$

Which means that the volume of the balloon will halve.

Learn more about ideal gases:

brainly.com/question/9321544

brainly.com/question/7316997

brainly.com/question/3658563

#LearnwithBrainly

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When you whisper, you produce a 10-dB sound.

lora16 [44]

All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
For c, 10 times 8 is 80 so it would be 80-dB

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4 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$

The force on the particle $q_1$ is $-180.38\ \text{N}$ .

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3 years ago

What type of climate would be near cold ocean currents? PLEASE HELP QUICK!!

olya-2409 [2.1K]

Answer:

hot humid with lots of rain.

Explanation:

ocean currents act as conveyer belts of warm and cold water sending heat to the polar regions and helping the tropical areas cool off, thus influencing both weather and climate. the tropics are particularly rainy because heat absorption , and thus ocean evaporation, is highest.

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4 years ago

Establish the dimension of density

mario62 [17]

Dimensional Formula of Density
The dimensional formula of density is given by,

[M1 L-3 T0]
Where,

M = Mass
L = Length
T = Time
Derivation

Density (ρ) = Mass × Volume-1 . . . . . (1)

The dimensional formula of volume = [M0 L3 T0] . . . . (2)

And, the dimensional formula of mass = [M1 L0 T0] . . . . (3)

On substituting equation (2) and (3) in equation (1) we get,

Density = Mass × Volume-1

Or, (ρ) = [M1 L0 T0] × [M0 L3 T0]-1 = [M1 L-3 T0]

Therefore, density is dimensionally represented as [M1 L-3 T0].

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