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3 years ago

So I don’t really get it can you also explain what’s the difference between radiation,conduction, and convection?

Physics

1 answer:

Slav-nsk [51]3 years ago

5 0

Answer:

I think its radiation

Explanation:

Conduction is the transfer of heat through solids (A)

Convection is the transfer of heat through liquids or gasses (B)

Radiation is the transfer of heat through em waves (C)

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If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu

Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

$\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}$ .

Volume of water:

$V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}$ .

Mass of water:

$m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}$ .

Amount of heat that the 15 L water absorbed:

$Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}$ .

What's the mass of the hot steel tool?

The specific heat of carbon steel is $0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}$ .

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

$Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}$ .

$\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}$ .

$m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}$ .

4 0

3 years ago

Un avión vuela a una velocidad de 900 km/h. Si tarda en viajar desde Canarias hasta la península 180 s ¿qué distancia recorre en

wel

Answer:

El avión recorrió 45 km en los 180 s.

Explanation:

La relación entre velocidad, distancia y tiempo se da de la siguiente manera;

$Velocidad= \dfrac{Distancia}{Hora}$

Por lo cual los parámetros dados son los siguientes;

Velocidad = 900 km/h = 250 m / s

Tiempo = 180 s

Estamos obligados a calcular la distancia recorrida

De la ecuación para la velocidad dada arriba, tenemos;

Distancia recorrida = Velocidad pf viaje × Tiempo de viaje

Distancia recorrida = 900 km/h × 180 s = 900

Distancia recorrida = 900 km/h × 1 h/60 min × 1 min/60 s × 180 s = 45 km

Por lo tanto, el avión viajó 45 km en 180 s.

8 0

4 years ago

When a net force of 17.0 newtons is applied to a dictionary placed on a frictionless table, it accelerates by 3.75 meters/second

4vir4ik [10]

Force = (mass) x (acceleration) (Newton's second law of motion)

Divide both sides of the equation by 'acceleration', and you have

Mass = (force) / (acceleration)

Mass = 17 newtons / 3.75 meters per second-sqrd = 4.533 kilograms (rounded)

8 0

3 years ago

Which statement describes the atomic number?<br>

Vadim26 [7]

Answer:The term atomic number, conventionally denoted by the symbol Z, indicates number of protons present in the nucleus of an atom, which is also equal to the number of electrons in an uncharged atom. The number of neutrons is represented by the neutron number (N)

Explanation:

6 0

3 years ago

You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c

Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

Since the block slides across the floor at constant speed, this means that it's not accelerated.
According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

$F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N (1)$

In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

$F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N (2)$

⇒ 169 N + Fn = Fg = 216 N (3)

This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
Fn = 216 N - 169 N = 47 N (4)

6 0

3 years ago