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1 year ago

What thermodynamic property describes the relationship of an equilibrium to temperature?

1 answer:

4 0

The thermodynamic property that describes the relationship between equilibrium and temperature is Thermodynamic equilibrium

Thermodynamics is a science that focuses on explaining chemically and physically phenomenons related to:

Heat
Temperature
Energy

In thermodynamics, the thermodynamic equilibrium occurs if the temperature or heat is in balance, and therefore it will not change in regular conditions. For example:

If a liquid has the same temperature as its surroundings, the temperature is likely to remain stable
The temperature of a liquid or solid does not change unless there are specific conditions
According to this, thermodynamic equilibrium describes the relation of equilibrium to temperature

Learn more in: brainly.com/question/22447172

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‏A 50 - N x m torque acts on a wheel with a moment of inertia 150 kg x m² . If the wheel starts from rest , how long will it tak

denis-greek [22]

Answer:

t = 6.17 s

Explanation:

For a 1 revolution movement, $\triangle \theta = 2\pi$

Torque, $\tau = 50 Nm$

Moment of Inertia, $I = 150 kg m^2$

If the wheel starts from rest, $w_{0} = 0 rad/s$

The angular displacement of the wheel can be given by the formula:

$\triangle \theta = \omega_0 t + 0.5 \alpha t^2$ ................(1)

Where $\alpha$ is the angular acceleration

$\tau = I \alpha\\\alpha = \frac{\tau}{I} \\\alpha = 50/150\\\alpha = 0.33 rad/s^2$

To get t, put all necessary parameters into equation (1)

$2\pi = 0(t) + 0.5(0.33)t^2\\2\pi =0.5(0.33)t^2\\t^2 = \frac{4 \pi}{0.33} \\t^2 = 38.08\\t = 6.17 s$

3 0

3 years ago

A 70-kg ice skater is standing still on the ice, when a friend tosses him a 10-kg ball that is traveling at +8 m/s. If he catche

Misha Larkins [42]

Answer:+1.25 m/s

Explanation:

Given

mass of ice skater M=70 kg

mass of ball m=10 kg

the initial velocity of the ball $u_1=+8\ m/s$

Conserving linear momentum

$M\times0+m\timesu_1=(M+m)v\quad \quad [v=\text{combined velocity of skater and ball}]$

$v=\dfrac{10\times10}{80}=+1.25\ m/s$

Therefore the velocity of the person holding the ball is 1.25 m/s

This collision represents the perfectly inelastic collision where particles stick together after the collision.

7 0

3 years ago

What is one way that microorganisms can be beneficial to humans

Anna11 [10]

"Some microbes<span> are used for medicinal production. One of the most important groups of medicines, antibiotics, is produced by fungi and bacteria. The name antibiotics means 'against life'. It is appropriate, because they attack bacteria and other unicellular organisms that are pathogenic for </span>humans<span>."

hope this helps you!</span>

3 0

3 years ago

Is there ever a situation where an ant will have more momentum than an elephant? Explain why or why not? Question 2 options: Yes

sergiy2304 [10]

An ant can have more momentum than an elephant when the elephant is standing still.

Answer: A

Explanation

The momentum is the quantification of the movement done by an object.

It is found to be dependent on the mass of the object and the velocity with which it is moving.

In the present case, the ant has negligible mass compared to elephant so the momentum can be more for ant only when the velocity with which the elephant is moving tends to be zero.

As the velocity of elephant will be zero, the momentum of elephant will be zero so in this criteria, the moving ant will be having more momentum compared to elephant with zero velocity.

So an elephant with zero velocity means the elephant is standing still.

Thus, the condition in which the ant will be having more momentum compared to elephant is when the elephant stands still.

6 0

2 years ago

Read 2 more answers

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel

insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t² (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t² (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t² (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

$p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125$

Since p = secθ

secθ = 1.95625 or secθ = -0.51125

cosθ = 1/1.95625 or cosθ = 1/-0.51125

cosθ = 0.5112 or cosθ = -1.9956

Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.

So, cosθ = 0.5112

θ = cos⁻¹(0.5112)

θ = 59.26°

So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.

5 0

3 years ago