Which body sense (choose only 1) do you think is most important? Explain why and provide at least 2

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 563 K, and r

Schach [20]

Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= $\frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}$

$= \frac{-1760}{563} + \frac{1760}{354}$

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

8 0

3 years ago

Could someone help me with this question please?

BabaBlast [244]

You are correct...amplitude will be the answer

3 0

3 years ago

Read 2 more answers

A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When

natita [175]

Answer:

a) $\mu_{k} = 0.704$ , b) $R = 0.312\,m$

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s$

$\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}$

$\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}$

$\mu_{k} = 0.704$

b) The speed of the block is determined by using the Principle of Energy Conservation:

$\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$v = x\cdot \sqrt{\frac{k}{m} }$

$v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }$

$v \approx 9.829\,\frac{m}{s}$

The radius of the circular loop is:

$\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}$

$\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}$

$R = 0.312\,m$

5 0

3 years ago

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the

aivan3 [116]

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^{\circ}$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$