Joule’s law is a linear relationship, that is, the more heat you provide, the greater the temperature change. However, in the pr

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.

The height measures 69.68 m

Explanation:

Given data

Before striking the ground =  46 % of the total distance

To establish

the height

Solution

We know here acceleration and displacement, which is

d = (0.5)gt²..............1

Here d is the distance, g is the acceleration, and t is time

So, when an object falls it will be

h = 4.9 t²....................2

For the first part of the inquiry

The falling objects account for

54 % of the total distance

0.54 h = 4.9 (t-1)²...................3

Thus,

Now we possess two equations with unknown variables

We can equate both equations

The first equation already solves for h

Substituting h in the second equation allows us to find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

We choose here 3.771 s since 0.576 s is negligible; the distance covered in the last second before it impacts the ground is 46 % of the entire fall.

Thus, selecting t = 3.771 s

Then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

Thus, the height is 69.68 m

a) 0.13*τ; b) 2.08*τ. To describe the discharging process of a capacitor through a resistor, consider the following: Q(t) = Qo * exp(-t/τ) to signify a loss of 1/8 of its charge. In this scenario, Q(t) = 7/8 * Qo = 7/8 * exp(-t/τ). By rearranging, we have ln(7/8)*τ = -t, thus t = -ln(7/8)*τ = 0.13. For a loss of 7/8 of its charge, we use Q(t) = 1/7 * Qo * exp(-t/τ), leading to t = -ln(1/8)*τ = 2.08.

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

                                   (1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

                        (10-5/12) × (21) = 218.75 meters.