All categories

2 years ago

A 12.7 L box is filled with 6.0 moles of Argon gas and cooled to a temperature of 210 K. What is the pressure?

Physics

1 answer:

sp2606 [1]2 years ago

3 0

Answer:

Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

Explanation:

You might be interested in

A 0.75 kilogram apple is thrown upward from the ground. The apple reaches a height of 5.5m, what is the beginning gravitational

MaRussiya [10]

Answer:

U(0.0m) = 0J

U(5.5m) = 40.42 J

Explanation:

The gravitational potential energy is given by the following formula:

$U=mgh$

h: height

m: mass of the apple = 0.75kg

g: gravitational acceleration = 9.8m/s^2

When the apple is at 5.5m from the ground the gravitational potential energy is:

$U=(0.75kg)(9.8m/s^2)(5.5m)=40.42\ J$

when the apple is on the ground you have:

$U=mg(0m)=0\ J$

6 0

3 years ago

Hair color is a trait that’s controlled by many genes working together. Each gene contributes to this trait only in a small prop

Katena32 [7]

Codominence i think, its both parents genes are used

3 0

2 years ago

A biconvex lens is formed by using a piece of plastic(n=1.70).

creativ13 [48]

Answer:

$f =17.15\ cm$

Explanation:

given,

refractive index of lens, n = 1.70

Radius of curvature of front surface. R₁ = 20 cm

Radius of curvature of the back surface, R₂ = 30 cm

focal length= ?

$\dfrac{1}{f}=(n-1)(\dfrac{1}{R_1}-\dfrac{1}{R_2})$

R₁ = +20 cm

R₂ = -30 cm

n = 1.70

$\dfrac{1}{f}=(1.70-1)(\dfrac{1}{20}-\dfrac{1}{-30})$

$\dfrac{1}{f}=0.70 \times 0.0833$

$f = \dfrac{1}{0.7 \times 0.0833}$

$f =17.15\ cm$

the focal length of the lens is equal to 17.15 cm

3 0

3 years ago

A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the

crimeas [40]

Answer:

Explanation:

The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²

where Ф is the angle of inclination, R is the radius, k is the radius of gyration.

The potential energy of the system is given as ; PE = mgh

The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.

The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2

Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

I = mR²

The moment of inertia of the ring is maximum and therefore reaches the bottom last.

7 0

2 years ago

At the end of cylindrical rod of length l = 1 m and mass M = 1 kg rotating horizontaly along the vertical axis in its center wit

matrenka [14]

Answer:

w = 0.943 rad / s

Explanation:

For this problem we can use the law of conservation of angular momentum

Starting point. With the mouse in the center

L₀ = I w₀

Where The moment of inertia (I) of a rod that rotates at one end is

I = 1/3 M L²

Final point. When the mouse is at the end of the rod

$L_{f}$ = I w + m L² w

As the system is formed by the rod and the mouse, the forces during the movement are internal, therefore the angular momentum is conserved

L₀ = L_{f}

I w₀ = (I + m L²) w

w = I / I + m L²) w₀

We substitute the moment of inertia

w = 1/3 M L² / (1/3 M + m) L² w₀

w = 1 / 3M / (M / 3 + m) w₀

We substitute the values

w = 1/3 / (1/3 + 0.02) w₀

w = 0.943 w₀

To finish the calculation the initial angular velocity value is needed, if we assume that this value is w₀ = 1 rad / s

w = 0.943 rad / s

3 0

3 years ago