Answer: B.) 6
Explanation:
To answer this problem you get the number of students who attended Wednesday (18) and the number of students who attended Tuesday (12) and subtract 18 - 12 = 6
Answer = B.) 6
<span>The magnetic field does not continually spread outward from the wire.</span>
Answer: In this lab we wanted to know how motion can be described. So the hypothesis is if the starting height of a sloped racetrack is increased, then the speed at which a toy car travels along the track will increase because the toy car will have a greater acceleration. My prediction is that cars travel faster on higher tracts. So the heighten the track was intentionally manipulated. So it is the independent variable the speed of the car is the dependent variable. The speed at the first quarter checkpoint is 1.09 m/s. The speed at the second quarter checkpoint is 1.95 m/s. The speed at the third quarter checkpoint is 2.373.36 m/s. The speed at the finish line is 2.803.00 m/s. The average speed increases as the height increases.
The cars on the higher track travel farther than the cars on the lower track, in the same time.
This means that the cars on the higher track have a greater average speed than those on the lower track. This is demonstrated by the
slope of the higher track line being greater than the slope of the lower track line.
Explanation: put it in notes then send it to files to compress it to submit it.
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
Answer: 
Explanation:
We know the total diameter of the cell (assumed spherical) is:
Then its total radius 
On the other hand, we know the thickness of the cell wall is 
and its density is the same as water (
).
Since density is the relation between the mass 
and the volume 
:
The mass is: 
(1)
Now if we are talking about this cell as a thin spherical shell, its volume will be:
(2)
Where 
Then:
(3)
(4)
Substituting (4) in (1):
(5)
(6)
Knowing 
and 
:
