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lions [1.4K]
3 years ago
10

How does the science of heat transfer differ from the science of thermodynamics?

Physics
1 answer:
mariarad [96]3 years ago
3 0

Answer:

The science of thermodynamics deals with the fundamental laws that guide how physical processes occur in relation with the energy transfer. When a system or process changes from one state of equilibrium to another, thermodynamics is interested with the amount of heat transfer during the process. On the other hand, the science of heat transfer is simply about the rate of heat and temperature distribution inside a system at a particular point in time.

Explanation:

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A block of ice with mass 5.50 kg is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal
Yuki888 [10]

Answer:

A) 3.13 m/s

B) 5.34 N

C) W = 26.9 J

Explanation:

We are told that the position as a function of time is given by;

x(t) = αt² + βt³

Where;

α = 0.210 m/s² and β = 2.04×10^(−2) m/s³ = 0.0204 m/s³

Thus;

x(t) = 0.21t² + 0.0204t³

A) Velocity is gotten from the derivative of the displacement.

Thus;

v(t) = x'(t) = 2(0.21t) + 3(0.0204t²)

v(t) = 0.42t + 0.0612t²

v(4.5) = 0.42(4.5) + 0.0612(4.5)²

v(4.5) = 3.1293 m/s ≈ 3.13 m/s

B) acceleration is gotten from the derivative of the velocity

a(t) = v'(t) = 0.42 + 2(0.0612t)

a(4.5) = 0.42 + 2(0.0612 × 4.5)

a(4.5) = 0.9708 m/s²

Force = ma = 5.5 × 0.9708

F = 5.3394 N ≈ 5.34 N

C) Since no friction, work done is kinetic energy.

Thus;

W = ½mv²

W = ½ × 5.5 × 3.1293²

W = 26.9 J

6 0
3 years ago
Explain what a concentration gradient is and what it means for a molecule to diffuse down its
MArishka [77]

Answer:

Diffusing the gradient ensures that most of the molecules in high concentration zone will wind up in the previously low concentration by the spontaneous movement of small molecules.

Explanation:

A gradient of concentration is the difference between in concentration of one place / area substance to different area. Having a molecule flow down its concentration gradient means moving the molecules from hypotonic areas to the concentration hypertonic areas

Diffusing the gradient ensures that most of the molecules in high concentration zone will wind up in the previously low concentration by the spontaneous movement of small molecules.

7 0
3 years ago
Physics. Need help. Brainlieast answer for most/ all of the answers answered
Mumz [18]

<u>ALL of the following work assumes NO AIR RESISTANCE:</u>

1). an object moving under the influence of only gravity, and not in orbit;  its horizontal velocity is constant, and its vertical motion is accelerated downward at 9.8 m/s²

2). a parabola

3). Horizontal: velocity is constant, acceleration is zero. . . . Vertical: acceleration is 9.8 m/s² downward, velocity depends on whether it was launched, thrown up, thrown down, dropped, etc.

4). a). the one that was thrown horizontally; b). both  hit the ground at the same time; c). both hit the ground with the same vertical velocity

5). a). zero; b). zero; c). gravity ... 9.8 m/s² down; d). 3.06 seconds; e). 4.38 m/s; f). 30 m/s g). no; gravity has no effect on horizontal motion

6). a). 1.8 seconds;   b). 13.1 meters;   c). 17.6 m/s down;   d). 7.3 m/s; gravity has no effect on horizontal motion

7). 45 m/s

8). without air resistance, the ball is traveling horizontally at 13 km/hr, and it lands back in your hand

9). a). 4.49 m/s;  b). 29.7 m/s

10). 7.24 meters

11).  700 meters

12).  A). 103.7 meters ( ! she's in big trouble ! );     B).  17.5 meters

3 0
3 years ago
Which of the following is an example of a noninfectious disease?
Scorpion4ik [409]
D lung cancer is not infectious
3 0
2 years ago
Read 2 more answers
To win the game, a place kicker must kick a
Dafna11 [192]

Answer:

1.86 m

Explanation:

First, find the time it takes to travel the horizontal distance.  Given:

Δx = 52 m

v₀ = 26 m/s cos 31.5° ≈ 22.2 m/s

a = 0 m/s²

Find: t

Δx = v₀ t + ½ at²

52 m = (22.2 m/s) t + ½ (0 m/s²) t²

t = 2.35 s

Next, find the vertical displacement.  Given:

v₀ = 26 m/s sin 31.5° ≈ 13.6 m/s

a = -9.8 m/s²

t = 2.35 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (13.6 m/s) (2.35 s) + ½ (-9.8 m/s²) (2.35 s)²

Δy = 4.91 m

The distance between the ball and the crossbar is:

4.91 m − 3.05 m = 1.86 m

5 0
3 years ago
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