A circular saw blade with a radius of 0.145 m starts from rest and turns in a vertical plane with a constant angular acceleratio
n of 2.00 rev/s2. After the blade has turned through 155 revs, a small piece of the blade breaks loose from the top of the blade. After the piece breaks loose, it travels with a velocity that is initially horizontal and equal to the tangential velocity of the rim of the blade. The piece travels a vertical distance of 0.820 m to the floor.
How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?
How far does the piece travel horizontally, from where it broke off the blade until it strikes the floor?
1 answer:
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Answer:
Explanation:
From the question;
We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.
We are to calculate the following task, i.e. to determine the electric field at the distances:
a) at 4.75 cm
b) at 20.5 cm
c) at 125.0 cm
Given that:
the charge (q) = 33.3 nC/m
= 33.3 × 10⁻⁹ c/m
radius of rod = 5.75 cm
a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.
Then, the electric field will be zero.
b) The electric field formula
E = 1461.95 N/C
c) The electric field E is calculated as:
E = 239.76 N/C