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umka2103 [35]
3 years ago
14

A group of air particles in a balloon are spread far apart and moving rapidly around inside the balloon. The balloon is placed i

n a refrigerator for 30 minutes. What would you expect to happen to the appearance of the balloon? Explain your answer in terms of particle movement and their change in temperature. The balloon is removed from the refrigerator and placed on the kitchen counter. What would you expect to happen to the appearance of the balloon? Explain your answer in terms of particle movement and their change in temperature.
Physics
1 answer:
Slav-nsk [51]3 years ago
6 0

Answer:

1) The balloon will shrink

2) The balloon will grow bigger and regain its shape

Explanation:

By the kinetic theory of gases, we have;

v_{rms} = \sqrt{\dfrac{3 \cdot R \cdot T}{MW} }

P = \dfrac{n \cdot MW \cdot vrms^2}{3 \cdot V}

Where;

v_{rms} = RMS Speed

R = The universal gas constant

T = The temperature

MW = The Molecular weight

n = The number of moles

V = The volume

P = The pressure

1) When the balloon is placed in the refrigerator for  minutes, the temperature inside the balloon will drop and RMS speed will reduce, however due to the elastic wall of the balloon, the pressure will remain the same therefore the volume  will reduce and the balloon will shrink

2) When the balloon is placed on the kitchen counter, the temperature inside the balloon will rise leading to the increase in the RMS speed which will raise the pressure effect on the wall  of the balloon as the balloon is made of elastic material, as the pressure rises, the wall expands to keep the pressure constant and the volume occupied by the air particles will increase, the balloon will grow bigger and regain its shape.

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Trees. Every time the wind blows there is a wave of motion which is movement

6 0
3 years ago
Read 2 more answers
You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find
ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

(c) μs = 0.5

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

<u>ω = 1.57 rad/s </u>

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

<u>ac = 4.92 m/s² </u>

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

<u>μs = 0.5</u>

7 0
2 years ago
A boy whirls a stone in a horizontal circle of radius 1.1 m and at height 2.1 m above ground level. The string breaks, and the s
DedPeter [7]

Answer:

212.8 m/s^{2}

Explanation:

Time taken by stone to cover horizontal distance

t=\sqrt{\frac {2h}{g}} where t is time, h is height of whirling the stone in horizontal circle, g is gravitational constant, Substituting h for 2.1 m and g for 9.81

t=\sqrt{\frac {2*2.1}{9.81}}= 0.654654 seconds

t=0.65 s

Velocity, v= distance/time

v=10/0.65= 15.27525 m/s

v=15.3 m/s

a=\frac {v^{2}}{r} where r is radius of circle, substituting r with 1.1m

a=\frac {15.3^{2}}{1.1}

a=212.8 m/s^{2}

Therefore, centripetal acceleration is 212.8 m/s^{2}

8 0
3 years ago
Block A, with a mass of 4.0 kg, is moving with a speed of 2.0 m/s while block B , with a mass of 8.0 kg, is moving in the opposi
Anon25 [30]
Consider velocity to the right as positive.

First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right

Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left

Total momentum of the system is
P = m₁v₁ + m₂v₂
   = 4*2 + 8*(-3)
  = -16 (kg-m)/s

Let v (m/s) be the velocity of the center of mass of the 2-block system.

Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s

Answer:
The center of mass is moving at 1.33 m/s to the left.
5 0
3 years ago
A thin rod of length 0.64 m and mass 120 g is suspended freely from one end. It is pulled to one side and then allowed to swing
valina [46]

Answer:

1. Kinetic Energy = 0.0161 Joules

2. Height = 0.0137m

Explanation:

Given

Length of Rod, l = 0.64m

Mass, m = 120g = 0.12kg

Angular speed, w = 1.40 rad/s

a.

Calculating the Rod's kinetic energy

This is calculated by

Kinetic Energy = ½Iw²

Where I = rotational inertia of the rod about an axis.

This is calculated as follows;

I = Icm + mh²

I = ImL² + m(L/2)²

I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²

I = 0.016384 kgm²

By substituton

KE = ½Iw² becomes

KE = ½ * 0.016384 * 1.40²

KE = 0.01605632J

KE = 0.0161 Joules

2. Using the total conservation of momentum;

K + U = Kf + V

Where K = Initial Kinetic Energy of the rod at lowest point.

U = Initial gravitational potential energy of the rod at lowest point

Kf = Final Kinetic Energy of the rod at maximum height = 0 J

V = Final gravitational potential energy of the rod at maximum height

So, K + U = Kf + V become

K + U = 0 + V

K + U = V

K = V - U = mgh

substitute 0.01605632J for K

0.01605632J = mgh

h = 0.01605632J/mg

h = 0.01605632J/(0.12 * 9.8)

h = 0.013653333333333

h = 0.0137m

4 0
3 years ago
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