the answer is 1a as rearrange gives I = v divided by r
Answer:
Object 3 has greatest acceleration.
Explanation:
Objects Mass Force
1 10 kg 4 N
2 100 grams 20 N
3 10 grams 4 N
4 1 kg 20 N
Acceleration of object 1,

Acceleration of object 2,

Acceleration of object 3,

Acceleration of object 4,

It is clear that the acceleration of object 3 is
and it is greatest of all. So, the correct option is (3).
Answer:
Leak 1 = 3.43 m/s
Leak 2 = 2.42 m/s
Explanation:
Given that the top of the boot is 0.3 m higher than the leaks.
Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2
From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.
Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2
Find the attached files for the solution and figure
Answer:

Explanation:
The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.
The above formula can be rewritten as follows

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.
If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.
Therefore the electric flux can be written as a function of time
