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Svetllana [295]
2 years ago
8

The alpha particles leave visible tracks in the cloud chamber because

Physics
1 answer:
julia-pushkina [17]2 years ago
6 0
<h2>Answer: Ionization </h2>

The inner atmosphere of a <u>cloud chamber</u> is composed of an easily ionizable gas, this means that little energy is required to extract an electron from an atom. <u>This gas is maintained in the supercooling state, so that a minimum disturbance is enough to condense it</u> in the same way as the water is frozen.

<h2>Then, when a charged particle with enough energy interacts with this gas, it <u>ionizes</u> it. </h2>

This is how alpha particles are able to ionize some atoms of the gas contained inside the chamber when they cross the cloud chamber.

These ionized atoms increase the surface tension of the gas around it allowing it to immediately congregate and condense, making it easily distinguishable inside the chamber like a <u>small cloud</u>. In this way, it is perfectly observable the path the individual particles have traveled, simply by observing the cloud traces left in the condensed gas.

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Select the correct expression that gives the block's acceleration at a displacement x from the equilibrium position. Note that x
skad [1K]

Answer:

d. a=−k/mx

Explanation:

To know what is the correct expression for the acceleration you take into account the second Newton law, that is:

F=ma ( 1 )

next, you equal the expression ( 1 ) to the force in a mass-string system, that is F=-kx.

ma=-kx\\\\a=-\frac{kx}{m}

hence, the acceleration is:

d. a=−kmx

8 0
2 years ago
Read 2 more answers
What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resti
Nataliya [291]

Archimedes principle states that

 

F1 / A1 = F2 / A2

F2 = (A2 / A1) * F1

 

Also, formula for the force is F = mg. Formula for the area of the cylinder is A = πr^2, therefore we get

 

F2 = (πr2^2 / πr1^2) * mg

 

Since the diameter of the cylinders are 2 cm and 24 cm, r1 = 12 and r2 = 1.

 

Substituting the values to the derived equation, we get

 

F2 = (π 1^2 / π 12^2) * 2400 * 9.8

F2 = 163.3333 N

 

 

<span> </span>

6 0
2 years ago
You are using a lightweight rope to pull a sled along level ground. The sled weighs 485 N, the coefficient of kinetic friction b
Bogdan [553]

Answer:

N=459.01N

Explanation:

According to Newton's first law:

N+F_y-W=0

The component of the force on the y-axis can be obtained through the Pythagorean Theorem. This is because the components are the cathetus of a right triangle and its hypotenuse is the magnitude of the force:

sin12^\circ=\frac{F_y}{F}\\F_y=Fsin12^\circ

Replacing and solving for N:

N=W-Fsin12^\circ\\N=485N-(125N)sin12^\circ\\N=459.01N

5 0
2 years ago
Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the s
frosja888 [35]

Answer:

C) True. S increases with time, v₁ = gt and v₂ = g (t-t₀)  we see that for the same t v₁> v₂

Explanation:

You have several statements and we must select which ones are correct. The best way to do this is to raise the problem.

Let's use the vertical launch equation. The positive sign because they indicate that the felt downward is taken as an opponent.

Stone 1

    y₁ = v₀₁ t + ½ g t²

    y₁ = 0 + ½ g t²

Rock2

It comes out a little later, let's say a second later, we can use the same stopwatch

     t ’= (t-t₀)

    y₂ = v₀₂ t ’+ ½ g t’²

    y₂ = 0 + ½ g (t-t₀)²

    y₂ = + ½ g (t-t₀)²

Let's calculate the distance between the two rocks, it should be clear that this equation is valid only for t> = to

    S = y₁ -y₂

    S = ½ g t²– ½ g (t-t₀)²

    S = ½ g [t² - (t²- 2 t to + to²)]  

    S = ½ g (2 t t₀ - t₀²)

    S = ½ g t₀ (2 t -t₀)

This is the separation of the two bodies as time passes, the amount outside the Parentheses is constant.

For t <to.  The rock y has not left and the distance increases

For t> = to.  the ratio (2t/to-1)> 1 therefore the distance increases as time

passes

Now we can analyze the different statements

A) false. The difference in height increases over time

B) False S increases

C) Certain s increases with time, v₁ = gt and V₂ = g (t-t₀) we see that for the same t   v₁> v₂

3 0
2 years ago
A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an
Nadya [2.5K]
Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm

Answer: 4.5 cm
7 0
2 years ago
Read 2 more answers
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