Response:
AB = 100 km; BC = 80 km; AC = 180 km
Time of arrival = 11:30
Reasoning:
1. Distance from A to B
(a) Duration of travel
Duration = 10:00 - 8:00 = 2.00 hours
(b) Distance
Distance = speed × time = 50 km/h × 2.00 h = 100 km
2. Distance from B to C
Distance = 80 km/h × 1 h = 80 km
3. Summary of Distances
AB = 100 km
BC = 80 km
AC = 180 km
4. Time of Arrival
Departure from A = 08:00
Travel duration to B = 2:00
Arrival at B = 10:00
Waiting time at B = 0:30
Departure from B = 10:30
Travel duration to C = 1:00
Arrival at C = 11:30
The required duration is 16.1 minutes. To determine the heat needed to raise the temperature, we must calculate the following amounts, where Q represents the required heat, m stands for mass, V represents the volume, C signifies specific heat, and ΔT indicates temperature change. After substituting the provided values into the formula and calculating, the next step is determining the required time based on the formula t = Q/P, where P is given as 1500 W. Ultimately, we find that the time needed is 16.1 minutes.
Answer:
The temperature of the cooler object was nearly at room temperature. As a result, the system underwent minimal change
Explanation:
In a closed system with two objects at varying temperatures, heat energy typically flows from the hotter object to the cooler one. This transfer is more pronounced when there is a significant temperature disparity between the objects. Conversely, if the temperature difference is minor or negligible, the resulting change will be minimal.
According to Snell’s Law:
Where: is the index of refraction of the first medium (glass), and is the index of refraction of the second medium (ice). The angle of incidence and the angle of refraction are represented by and. The refractive index quantifies the speed of light in a medium. The critical angle is identified as the angle at which total internal reflection occurs, meaning no light passes through into another medium. This phenomenon happens only when the light is transitioning from a medium with a higher index of refraction to one with a lower index of refraction.
