Think of one example where someone would need to calculate the net force on a person at the waater park
1 answer:
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Answer:
Temperature on the inside ofthe box
Explanation:
The power of the light bulb is the rate of heat conduction of the bulb,
The thickness of the wall, L = 1.2 cm = 0.012m
Length of the cube's side, x = 20cm = 0.2 m
The area of the cubical box, A = 6x²
A = 6 * 0.2² = 6 * 0.04
A = 0.24 m²
Temperature of the surrounding,
Temperature of the inside of the box,
Coefficient of thermal conductivity, k = 0.8 W/m-K
The formula for the rate of heat conduction is given by:
The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
