To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

Here,
q = Charge
t = Time
R = Resistance
C = Capacitance
When the charge reach its half value it has passed 10ms, then the equation is,




We know that RC is equal to the time constant, then

Therefore the time constant for the process is about 14ms
Answer:
3rd picture straight line going up right
Explanation:
3rd picture
Kick | Distance (M) | Time (s) | Average Speed
1. 55. 5.0. 11
2. 50. 5.0. 10
3. 20. 3.0. 10
The slope of the road can be given as the ratio of the change in vertical
distance per unit change in horizontal distance.
- The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.
Reasons:
Width of the truck = 2.4 meters
Height of the truck = 4.0 meters
Height of the center of gravity = 2.2 meters
Required:
The allowable steepness of the slope the truck can be parked without tipping over.
Solution:
Let, <em>C</em> represent the Center of Gravity, CG
At the tipping point, the angle of elevation of the slope = θ
Where;

The steepness of the slope is therefore;

Where;
= Half the width of the truck =
= 1.2 m
= The elevation of the center of gravity above the ground = 2.2 m



The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.
Learn more here:
brainly.com/question/20793607
When an object's atoms move faster, its thermal energy increases and the object becomes warmer.