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Salsk061 [2.6K]
2 years ago
8

When a balloon is deflating, why does air leave the balloon?

Physics
1 answer:
Gwar [14]2 years ago
6 0

Answer:

<em>When a balloon deflates air moves out of the balloon </em><em>because the pressure inside the balloon is higher than the pressure outside the balloon.</em>

Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.

This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.

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A 6.0-cm-diameter horizontal pipe gradually narrows to 4.5 cm. When water flows through this pipe at a certain rate, the gauge p
Rufina [12.5K]

Answer:

0.0072 m³/s

Explanation:

Using Bernoulli's law

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂ since the pipe is horizontal

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

flow rate is constant

A₁v₁ = A₂v₂

A₁ = πr₁² = π (0.06/2)² = 0.0028278 m²

A₂ = πr₂² = π (0.0225)² = 0.00159 m²

v₁  = (A₂ / A₁)v₂

v₁  = (0.00159 m²/ 0.0028278  m²) v₂ = 0.562  v₂

substitute v₁  into the Bernoulli's equation

1/2ρv₂² - 1/2ρv₁² = P₁ - P₂

500 ( 1 - 0.3161 ) v₂²  = (31.0 - 24 ) × 10³ Pa

341.924 v₂² = 7000

v₂² = 20.472

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volume follow rate = 0.00159 m² ×  4.525 m/s = 0.0072 m³/s

8 0
2 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
Phoenix [80]

Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
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