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seropon [69]
3 years ago
8

In un lento processo di riscaldamento di 200 g di H2O da 60 gradi a 100 gradi evaporano 10 g di H2O. Assumendo per il calore lat

ente di vaporizzazione dell'H2O il valore costante di 540 cal/g, stabilire , con un calcolo di prima approssimazione, il calore totale che si deve fornire all'H2O per eseguire il riscaldamento
soluzione: Q=13 000 circa cal
Physics
1 answer:
Lyrx [107]3 years ago
7 0

https://www.khanacademy.org/science/physics/thermodynamics/specific-heat-and-heat-transfer/v/specific-heat-and-latent-leat-of-fusion-and-vaporization-2

https://www.khanacademy.org/science/physics/thermodynamics/specific-heat-and-heat-transfer/v/specific-heat-and-latent-leat-of-fusion-and-vaporization-2

This link might help

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Describe each of Newton’s Laws of Motion in ice skating. What can you design/develop to improve ice skating?
denis23 [38]

Newton's three laws of motion can be used to describe the motion of the ice skating.

<h3>Newton's first law of motion</h3>

Newton's first law of motion states that an object at rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.

  • Based on this law, once the ice skating starts, it will continue endlessly unless external force stops it.

<h3>Newton's second law of motion</h3>

Newton's second law of motion states that the force applied to an object is directly proportional to the product of mass and acceleration of an object.

  • Based on this law, the force applied to the ice skating is equal to the product of mass and acceleration of the ice skating.

<h3>Newton's third law of motion</h3>

This law states that action and reaction are equal and opposite.

  • Based on this law, the force applied to the ice skating is equal in magnitude to the reaction of ice.

Learn more about Newton's law here: brainly.com/question/3999427

3 0
3 years ago
Read 2 more answers
A 1 530-kg automobile has a wheel base (the distance between the axles) of 2.70 m. The automobile's center of mass is on the cen
NeTakaya

Answer:

Force on front axle = 6392.85 N

Force on rear axle = 8616.45 N

Explanation:

As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels

Now we know that

F_1 + F_2 = W

F_1 + F_2 = (1530\times 9.81)

F_1 + F_2 = 15009.3 N

now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle

so we can write torque balance about its center of mass

F_1(1.15) = F_2(2.70 - 1.15)

F_1 = 1.35 F_2

now from above equation

F_2 + 1.35F_2 = 15009.3

now we have

F_2 = 6392.85 N

now the other force is given as

F_1 = 8616.45 N

4 0
3 years ago
[]Answer the question below[]
marysya [2.9K]
Answer:

The answer is D. density.
8 0
2 years ago
As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax
AysviL [449]

Answer:

I=2.766\ kg.m^2

Explanation:

We have:

diameter of the wheel, d=0.88\ m

weight of the wheel, w_w=280\ N

mass of hanging object to the wheel, m_o=6.32\ kg

speed of the hanging mass after the descend, v_o=4\ m.s^{-1}

height of descend, h=2.5\ m

(a)

moment of inertia of wheel about its central axis:

I=\frac{1}{2} m.r^2

I=\frac{1}{2} \frac{w_w}{g}.r^2

I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2

I=2.766\ kg.m^2

3 0
3 years ago
Two blocks connected by a rope of negligible mass are being dragged by a horizontal force (see figure below). Suppose F = 65.0 N
vovangra [49]
Refer to the diagram shown below.

g = 9.8 m/s², and air resistance is ignored.

For mass m₁:
The normal reaction is m₁g.
The resisting force is R₁ = μm₁g.

For mass m₂:
The normal reaction is m₂g.
The resisting force is R₂ = μm₂g.

Let a =  the acceleration of the system.
Then
(m₁ + m₂)a = F - (R₁ + R₂)
(14+26 kg)*(a m/s²) = (65 N) - 0.098*(9.8 m/s²)*(14+26 kg)
40a = 65 - 38.416 = 26.584
a = 0.6646 m/s²

Answer:  0.665 m/s²  (nearest thousandth)

7 0
3 years ago
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