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3 years ago

Which is not an assumption about particles in a gas according to the kinetic theory?

1 answer:

4 0

Let us evaluate the given assumptions according to the kinetic theory for an ideal gas.

a.
The motion of one particle is unaffected by other particles unless the particles collide.
TRUE. The particles are in random motion unless they collide.

b.
The forces of attraction among particles keep the particles close together.
FALSE. No forces act between particles except during collision.

c.
Under ordinary conditions, forces of attraction between particles can be ignored.
TRUE.

Answer: Statement b is false because it is not an assumption.

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A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o

aniked [119]

Correct question is;

A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers a damping force numerically equal to √2 times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 7 ft/s. (Use g = 32 ft/s²)

Answer:

x(t) = 7te^(-2t√2)

Explanation:

We are given;

Weight; W = 8 lbs

mass; m = W/g

g = 32 ft/s²

Thus;

m = 8/32

m = ¼ slugs

From Newton's second law we can write the equation as;

m(d²x/dt²) = -kx - β(dx/dt)

Rearranging this, we have;

(d²x/dt²) + (β/m)(dx/dt) + (k/m)x = 0

Where;

β is damping constant = √2

k is spring constant = W/s

Where s = 8ft - 4ft = 4ft

k = 8/4

k = 2

Thus,we now have;

(d²x/dt²) + (√2/(¼))(dx/dt) + (2/(¼))x = 0

>> (d²x/dt²) + (4√2)dx/dt + 8x = 0

The auxiliary equation of this is;

m² + (4√2)m + 8 = 0

Using quadratic formula, we have;

m1 = m2 = -2√2

The general solution will be gotten from;

x_t = c1•e^(mt) + c2•t•e^(mt)

Plugging in the relevant values gives;

x_t = c1•e^(mt) + c2•t•e^(mt)

At initial condition of t = 0, x_t = 0 and thus; c1 = 0

Also at initial condition of t = 0, x'(0) = 7 and thus;

Since c1 = 0, then c2 = 7

Thus,equation of motion is;

x(t) = 7te^(-2t√2)

8 0

3 years ago

All moving objects don’t have momentum A. True B. False

mestny [16]

Answer:

Explanation:

Some objects gain momentum.

4 0

2 years ago

Read 2 more answers

A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The d

ValentinkaMS [17]

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= - 4 m/s

8 0

3 years ago

The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ

prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

$\nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})$

= $(30)^{2} + 2(-0.25(412 - 0))$

= 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

y = $200 e^{\frac{x}{1000}}$

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

Now, second derivative of the train is calculated as follows.

$\frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}$

$\frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}$

Radius of curvature of the train is as follows.

$\rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}$

= $\frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}$

= 3808.96 m

Now, we will calculate the normal component of the train as follows.

$a_{n} = \frac{\nu^{2}_{B}}{\rho}$

= $\frac{(26.34)^{2}}{3808.96}$

= 0.1822 $m/s^{2}$

The magnitude of acceleration of train is calculated as follows.

a = $\sqrt{(a_{t})^{2} + (a_{n})^{2}}$

= $\sqrt{(-0.25)^{2} + (0.1822)^{2}}$

= $0.309 m/s^{2}$

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is $0.309 m/s^{2}$ .

6 0

3 years ago

Galaxies are separated by?

Inessa [10]

It is number c I think

8 0

3 years ago

Read 2 more answers