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galben [10]
3 years ago
5

As the temperature of a fluid decreases— A The number of inter-particle collisions decrease and random movement of particles inc

rease. B The number of inter-particle collisions and random movement of particles decrease. C The number of inter-particle collisions and random movement of particles increase. D The number of inter-particle collisions increase and random movement of particles decrease.
Physics
2 answers:
Alenkasestr [34]3 years ago
7 0

Answer:

Only option 'B' is correct

Explanation:

Other options are not totally correct. The molecular collision and random movement of particles are proportional so they both decrease in this condition. Thank you.

Alika [10]3 years ago
3 0

Answer:

Option B The number of inter-particle collisions and random movement of particles decrease.

Explanation:

According to the kinetic theory of gas,

The average kinetic energy of gas molecules are directly proportional to the temperature. This means that an increase in temperature will give the gas molecules more energy to move randomly and a decrease in the temperatures will also decrease the energy of the molecules hence decreasing their random movement. Also, as the temperature increases, and the gas molecules moves more randomly, they collide more with each other and as the temperature decrease, and their random movement also decrease, the collision of the gas molecules also decrease.

Now from the question given above,

As the temperature of a fluid decreases, the number of inter-particle collisions and random movement of particles decrease.

So, option B is the correct option.

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If the angular frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum acceleration
Nataly_w [17]

Answer:

When we double the angular velocity the maximum acceleration (a_{max}) will changes by a factor of 4.

Explanation:

Given the angular frequency (\omega) of the simple harmonic oscillator is doubled.

We need to find the change in the maximum acceleration of the oscillator.

a_{max}=A\omega^2

Now, according to the problem, the angular frequency (\omega) got doubled.

Let us plug \omega=2\times \omega. Then the maximum acceleration will be a_{max'}

a_{max}=A\omega^2

a_{max'}=A(2\times \omega)^2\\a_{max'}=A\times 4\omega\\a_{max'}=4A\omega

a_{max'}=4a_{max}

We can see, when we double the angular velocity the maximum acceleration will changes by a factor of 4.

6 0
2 years ago
Which measurement is a potential difference?<br> A.115J<br> B.115V<br> C.115N<br> D.115C
borishaifa [10]

Answer:

B

Explanation:

Potential difference has a SI Unit of Volt and its symbol is <em>V</em>. Hence answer is <u>B</u>.

A is wrong as it has the unit Joule <em>(J)</em> which is the SI unit for energy.

C is wrong as it has the unit Newton <em>(N)</em> which is the SI unit for force.

D is wrong as it has the unit Coulomb <em>(C)</em> which is the SI unit of charge.

5 0
3 years ago
What is the relationship between an object’s mass and its gravitational potential energy?.
antiseptic1488 [7]

Answer:

If we’re talking about objects on the Earth, the gravitational potential energy is given by:

Explanation:

PEg=mgh

so the energy is proportional to the mass ( m ), but also to the strength of the gravitational field ( g ), and the height ( h ) to which the mass is lifted.

5 0
2 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
The magnetic domains in a non-magnetized piece of iron are characterized by which orientation
mel-nik [20]
The unmagnetized pieces of iron would be randomly pointing to directions, this is true because although influenced with the magnetic domain, the direction of the unmagnetized iron field of attraction is not uniform or does not have preferred direction.
3 0
3 years ago
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