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Salsk061 [2.6K]
3 years ago
8

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

Physics
1 answer:
Gwar [14]3 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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Nastasia [14]
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3 0
3 years ago
The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3 kg a
katen-ka-za [31]

Answer:

Mg= 1.3418 kg*m/s

Explanation:

Mass of rifle

m_{r}=2.90 kg

Recoil velocity of rifle

v_{r}=1.95 m/s

Momentum of rifle

m*v_{r}=3.0kg*1.95\frac{m}{s}==5.85 \frac{kg*m}{s}

mass of bullet

mb=7.2x10^{-3} kg=0.0072 kg

velocity of bullet relative to muzzel

v_{b} = 601 m/s

velocity of bullet relative to earth= 601 - 1.95=599.05 m/s

momentum of bullet

m_{b}*m_{b}=7.2x10^{-3}kg*599.05\frac{m}{s}=4.3132 \frac{kg*m}{s}

the momentum of the propellant gases = momentum of rifle - momentum of bullet

the momentum of the propellant gases

Mg=5.655 -4.3132

Mg= 1.3418 kg*m/s

the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle is 1.3418 kgm/s

5 0
3 years ago
One molecule of bromine (Br2) and two molecules of potassium chloride (KCI) combine in a reaction. How many atoms are in the pro
Liula [17]

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Let's consider the balanced equation for the reaction between 1 molecule of bromine and 2 molecules of potassium chloride. This is a single replacement reaction.

Br₂ + 2 KCl ⇒ 2 KBr + Cl₂

We obtain as products, 2 molecules of potassium bromide and 1 molecule of chlorine.

  • 1 molecule of KBr has 2 atoms, so 2 molecules contribute with 4 atoms.
  • 1 molecule of Cl₂ has 2 atoms.
  • The 4 atoms from KBr and the 2 atoms from Cl₂ make a total of 6 atoms.

In the reaction between 1 molecule of bromine and 2 molecules of potassium chloride, there are six atoms in the products.

Learn more: brainly.com/question/21850455

5 0
3 years ago
Read 2 more answers
Which is an example of nuclear energy being converted into heat and light energy?
likoan [24]
The answer to the question is A 
6 0
3 years ago
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Find the net force FnetFnetF_net acting on the sled. Express your answer in terms of some or all of the variables mmm, sss, v1v1
kkurt [141]

Complete Question

The kinetic energy K of an object of mass m moving at a speed v is defined as . It seems reasonable to say that the speed of an object--and, therefore, its kinetic energy--can be changed by performing work on the object. In this problem, we will explore the mathematical relationship between the work done on an object and the change in the kinetic energy of that object.

Let us now consider the situation quantitatively. Let the mass of the sled be m and the magnitude of the net force acting on the sled be    The sled starts from rest.

Consider an interval of time during which the sled covers a distance s and the speed of the sled increases from v_1 to v_2. We will use this information to find the relationship between the work done by the net force (otherwise known as the net work) and the change in the kinetic energy of the sled.

Find the net force acting on the sled.

Express your answer in terms of some or all of the variables m,s,v_1, and v_2.

Answer:

The expression is  F_{net}   = \frac{1}{2s}  * m *  (v_2^2 -  v_1^2)

Explanation:

From the question we are told that

   The net force is F_{net}

    The  distance is  s

     The first velocity is  v_1

     The second velocity is  v_2

     The mass is  m

     

Generally the work energy theorem is mathematically represented as

       W =  F_{net} *  s

Also from the law energy conservation workdone is mathematically represented as

      W = \Delta  K

Here  \Delta K is the change in kinetic energy and this is mathematically represented as

      \Delta K  =  \frac{1}{2}  *  m  * \Delta v^2

So

         W  =  \frac{1}{2}  * m *  \Delta v^2

Here  

      \Delta v^2  =  v^2_2 - v^2_1

Hence

         W  =  \frac{1}{2}  * m *  (v_2^2 -  v_1^2)

So

       F_{net} *  s  = \frac{1}{2}  * m *  (v_2^2 -  v_1^2)

=>    F_{net}   = \frac{1}{2s}  * m *  (v_2^2 -  v_1^2)

4 0
3 years ago
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