In Hypothesis #1, the

The vector quantity that defines the distance and direction between two positions. It is a change in your position.

lilavasa [31]

I believe you're talking about displacement. It's a directional vector that depicts the movement of a point between two instances.

4 0

3 years ago

Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.

Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

$\eta = 1-\frac{T_C}{T_H}$

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

$T_C = 270^{\circ}C+273=543 K$

$T_H = 450^{\circ}C+273=723 K$

Therefore the maximum efficiency is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249$

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

$T_H = 270^{\circ}C+273=543 K$ is the high-temperature reservoir

$T_C = 150^{\circ}C+273=423 K$ is the low-temperature reservoir

If we apply again the formula of the efficiency

$\eta = 1-\frac{T_C}{T_H}$

The maximum efficiency of the second engine is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221$

8 0

3 years ago

Your cousin Jannik skis down a blue square ski slope, with an initial speed of 3.6 m/s. He travels 15 m down the mountain side b

fenix001 [56]

Answer: The loss of energy due to friction is equal to 1,253 J.

Explanation:

The problem tells us that the skier has an initial speed of 3.6 m/s, which means that his initial kinetic energy is as follows:

K₁ = 1/2 m v₁² = 1/2 . 58.0 Kg. (3.6)² (m/s)² = 376 J

After coming to a flat landing, his final speed is 7.8 m/s, so the final kinetic energy is as follows:

K₂ = 1/2 m v₂² = 1/2. 58.0 Kg. (7.8)² (m/s)² = 1,764 J

Now, when skying down the slope the increase in kinetic energy only can come from another type of energy, in this case, gravitational potential energy.

If we take the ground flat level as a Zero reference, the initial gravitational potential energy, can be written as follows, by definition:

U₁ = m.g. h (1)

Now, we don't know the value of the height h, but we know that the incline has a 18º angle above the horizontal, and that the distance travelled along the incline is 15 m.

By definition, the sinus of an angle, is equal to the proportion between the height and the hypotenuse , so we can write the following equation:

sin 18º = h / 15 m ⇒ h = 15 m. sin 18º = 4.6 m

Replacing in (1), we get:

U₁ = 58.0 Kg. 9.8 m/s². 4.6 m = 2,641 J

So, we can get the total initial mechanical energy, as follows:

E₁ = K₁ + U₁ = 376 J + 2,641 J = 3,017 J

After arriving to the flat zone, all potential energy has become in kinetic energy, even though not completely, due to the effect of friction.

This remaining kinetic energy can be written as follows:

E₂ = K₂ = 1,764 J

The difference E₂-E₁, is the loss of energy due to friction forces acting during the travel along the 15 m path, and is as follows:

ΔE= E₂ - E₁ = 1,764 J - 3,017 J = -1,253 J

8 0

3 years ago

A 3.00 kg cart on a frictionless track is pulled by a string so that it accelerates at 2.00 m/s/s. what is the tension in the st

AnnZ [28]

Maybe the picture helps. The blue block represents the cart with a mass of 3 kg. The person(black block) is pulling the cart to the right with a force F so that the acceleration a is 2 m/s². According to Newton's 2nd law: F = m*a.