A class of students performed the heat transfer experiment shown in the picture. In the experiment, three plastic dishes were pa
rtially filled with equal amounts of dark soil, light soil, and water. Thermometers were placed in each dish, and the dishes were placed under a heating lamp. The substances were left under the heating lamp for ten minutes, and their temperatures were observed and recorded. After recording the temperatures of the substances as they were heating, the students turned the lamp off and recorded the temperatures of the substances as they cooled for ten minutes.
Which method of heat transfer is illustrated in this experiment? Explain your answer
Which method of heat transfer is illustrated in this experiment? Explain your answer
1 answer:
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The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is .
The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.
The given data:
time, t = 1.1 s
initial speed, u = 1000 km/h =
final speed, v = 0 m/s
So we will be using the equation of motion, that is,
v = u + at
Hence , the deceleration of the rocket is .
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Answer:
La velocidad con la que se debe lanzar el trineo es aproximadamente 9.96 m/s
Explanation:
La masa del trineo, m = 2 kg
El ángulo de inclinación del trineo, θ = 30 °
El coeficiente de fricción, μ = 0,15
La altura a la que debe ascender el trineo, h = 4 m
La reacción normal del trineo en la rampa, N = m · g · cos (θ)
La fuerza de fricción = N × μ
Dónde;
g = Th aceleración debida a la gravedad ≈ 9.81 m/s²
∴ = 0.15 × 2 kg × 9.81 m/s² × cos (30°) ≈ 2.55 N
La longitud de la rampa que se mueve el trineo, l = h/(sin(θ))
∴ l = 4/(sin(30°)) = 8
La longitud de la rampa que se mueve el trineo, l = 8 m
El trabajo realizado sobre la fricción, =
× l
= 2.55 × 8 ≈ 20.4
El trabajo realizado sobre la fricción, ≈ 20.4 Julios
El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. = m·g·h
∴ P.E. = 2 kg × 9.81 m/s² × 4 m ≈ 78.48 Joules
El trabajo realizado para levantar el trineo a 4 metros de altura, P.E. ≈ 78.48 Joules
El trabajo total requerido para levantar el trineo 4 metros por la rampa, =
+ P.E.
= 20.4 J + 78.84 J ≈ 99.24 J
Energía cinética, K.E. = 1/2 × m × v²
La energía cinética necesaria para mover el trineo por la rampa, K.E. se da de la siguiente manera;
K.E. = 1/2 × 2 kg × v² ≈ 99.24 J
∴ v² ≈ 99.24 J/(1/2 × 2 kg)
La velocidad con la que se debe lanzar el trineo para que ascienda 4 metros por la rampa, v ≈ 9.96 m/s