3 years ago

Which, if any, of the heat engines violate(s) the second law of thermodynamics?

Physics

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

The second law states that the total entropy can never decrese over time for an isolated system

You might be interested in

Topics related to physics

Bezzdna [24]

Answer:

There are many topics related to physics such as :

Kinematics

Dynamics

Light

Sound

7 0

2 years ago

A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely i

blsea [12.9K]

Answer:

7066kg/m³

Explanation:

The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:

Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)

= 7066kg/m³

Explanation:

8 0

3 years ago

Read 2 more answers

Dianne's teacher has a chunk of dry ice for the class to look at. Dry ice is a kind of frozen gas—it is very cold. The teacher w

Gelneren [198K]

The answer is A. Hope this helps you with your work.

4 0

3 years ago

Read 2 more answers

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$ .

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0

3 years ago

A baseball pitcher throws a ball at 40 ms^-1. if the acceleration is approximately constant over a distance of 2 m, how large is

Reil [10]

We have the equation of motion $v^2=u^2+2as$ , where v i the final velocity, u is the initial velocity, a is the acceleration and s is the displacement

Here final velocity, v = 40m/s

Initial velocity, u = 0 m/s

Displacement s = 2 m

Substituting $40^2=0^2+2*a*2\\ \\ a=400m/s^2$

So the baseball pitcher accelerates at 400m/ $s^2$ to release a ball at 40 m/s.

4 0

3 years ago