Answer:
Explanation:
The food chopping appliances in this scenario should adapt use of digital image processing systems to capture the images after the chopping begins. Certain variables can be defined like food particles size, area etc. This will help deciding a threshold for the variable and then the comparator used in the system will(run by real time operating system) compare the threshold size with the digital image size of particles. If the size is found small then the speed of motor should reduce meaning we have chopped the food as per our need. In case the chopping is not done aptly, the same can be enhanced by increasing the speed of motor till the final outcome is reached
Answer:
i. -4m
ii. 20m
Explanation:
The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)
Total distance = 8m going east + 8m back to origin + 4m west = 20m
Answer:
26.82m/s
Explanation:
Given
Mass = m= 0.4kg
Initial Velocity = u = 0
Charge = 4.0E-5C
Distance= d = 0.5m
Object Charge = 2E-4C
First, we'll calculate the initial energy (E)
E = Potential Energy
PE = kQq / d
Where k = coulomb constant = 8.99E9Nm²/C²
Energy is then calculated by;
PE = 8.99E9 * 4E-5 * 2E-4 / 0.5
PE = 143.84J
Energy = Potential Energy = Kinetic Energy
K.E = ½mv² = 143.84J
½mv² = ½ * 0.40 * v² = 143.85
0.2v² = 143.85
v² = 143.85/0.2
v² = 719.25
v = √719.25
v = 26.81883666380777
v = 26.82m/s
Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge
Sorry to say but I know that t(e introduction is first and the coda is last
