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3 years ago

Can an argon atom undergo vibrational motion?

1 answer:

azamat3 years ago

5 0

No. the answer to the question if can an argon atom undergo vibrational motion is no. it can not even spin either. the argon atom, or the argon is a chemical element that is the third most abundant gas in the earth's atmosphere. it is ore than twice as abundance as water vapor. Thank you for this question.

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A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

3 years ago

What is the modern periodic table of elements?

Kamila [148]

C is the correct answer

3 0

2 years ago

Read 2 more answers

Un trineo de 20 kg descansa en la cima de una pendiente de 80 m de longitud y 30° de inclinación. Si µ = 0.2, ¿cuál es la veloci

Mariulka [41]

Answer:

v= 26.70 m/seg

Explanation: Ver anexo ( diagrama de cuerpo libre)

De acuerdo a la segunda ley de Newton

∑ F = m*a

∑ Fx = m* a(x) ∑ Fy = m* a(y)

También sabemos que el coeficiente de roce dinámico es:

μ = 0.2 = F(r)/N siendo N la fuerza normal.

Si descomponemos la fuerza P = mg = 20Kg* 9.8m/seg²

P = 196 [N] en sus componentes sobre los ejes x y y tenemos

Py = P* cos30 = 196* √3/2 = 98*√3

Px = P* sen30 = 196*1/2 = 98

La sumatoria sobre el eje y es :

∑ F(y) = m*a Py - N = 0 98*√3 = N ( no hay movimiento en la dirección y)

∑ F(x) = m*a P(x) - Fr = m*a

Fr = μ *N = 0.2* 98*√3

Fr = 19.6*√3 [N]

98 - 19.6*√3 = m*a

98 - 33.52 = m*a

a = (98 - 33.52 ) / 20

a = 3.22 m/seg²

Para calcular la velocidad del trineo al pié del plano, sabemos que al pié del plano el trineo ha recorrido 80 m, y que de cinemática

v² = v₀² + 2*a*d ( se pueden chequear unidades para ver la consistencia de la ecuación v y v₀ vienen dados en m/seg entonces v² y v₀² vienen en m²/seg², el producto de a (m/seg²) por la distancia d (m) resulta en m²/seg² entonces es consistente la relación

v² = 0 + 2*3.22*80 ( la velocidad inicial es cero)

v² = 515.2 m²/seg²

v = √515.2 m/seg

v= 26.70 m/seg

6 0

3 years ago

Two charged objects separated by some distance attract each other. If the charges on both objects are doubled with no change in

Serggg [28]

Answer:

(a) The force between them quadruples

Explanation:

According to coulomb's law, initial force between the two charged objects is given as;

$F_1=\frac{Kq_1q_2}{r^2}$

where;

k is coulomb's constant

q₁ is the charge on the first object

q₂ is the charge on the second object

r is the distance between the two objects

When the charges on both objects are doubled, then;

q₁ = 2q₁

q₂ = 2q₂

Force between the two charged objects will become

$F_2 = \frac{K2q_12q_2}{r^2} = \frac{4Kq_1q_2}{r^2} = 4(\frac{Kq_1q_2}{r^2}) = 4F_1$

Therefore, the force between them quadruples

4 0

3 years ago

Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba

mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

3 0

3 years ago

Read 2 more answers