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Solnce55 [7]
3 years ago
15

An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is t

he volume of the bubble when it reaches the surface? Assume that the temperature and the number of air molecules in the bubble remain constant during the ascent.
Physics
1 answer:
murzikaleks [220]3 years ago
4 0

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

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Answer:

true

Explanation:

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3 0
3 years ago
A 6 N and a 10 N force act on an object. The moment arm of the 6 N force is 0.2 m. If the 10 N force produces five times the tor
Levart [38]

Answer:

The moment arm is 0.6 m

Explanation:

Given that,

First force F_{1}=6\ N

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We need to calculate the moment arm

Using formula of torque

\tau=Force\times lever\ arm

So, Here,

\tau_{2}=5 \tau_{1}

We know that,

The torque is the product of the force and distance.

Put the value of torque in the equation

F_{2}\times d_{2}=5\times F_{1}\times r_{1}

r_{2}=\dfrac{5\times F_{1}\times r_{1}}{F_{2}}

Where, F_{1}=First force

F_{1}=First force

F_{2}=Second force

r_{1}= distance

Put the value into the formula

r_{2}=\dfrac{5\times6\times0.2}{10}

r_{2}=0.6\ m

Hence, The moment arm is 0.6 m

6 0
3 years ago
If an object is moving up, is it considered to have a positive or negative velocity?
grin007 [14]

Answer:

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7 0
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You will be colliding two carts in this lab. Select all the systems for which you predict the total momentum of the system will
baherus [9]

From the momentum conservation we know that the initial momentum is equal to the final momentum. The momentum in a singular way can be defined as the product between the mass and the velocity of an object. In the presented system, however, there are two objects, therefore the mass of both and the speed of both, before and after the collision must be taken into account. Mathematically we could describe this as

m_1u_1+m_2u_2 = m_1v_1+m_2v_2

Here,

m_{1,2} = Mass of each object

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From here we can realize that it is necessary to use the system on both cars to be able to predict what will happen either with their masses, or their speeds.

The correct answer is C.

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3 years ago
Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface be
miv72 [106K]

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

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h = 44.1 m

4 0
3 years ago
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