Complete Question
A very large sheet of a conductor carries a uniform charge density of
on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is 
Explanation:
From the question we are told that
The charge density is 
The position outside the surface is
Generally the electric field is mathematically represented as

Where
is the permitivity of free space with values 
substituting values


Answer:
The output power the weightlifter is 2916.67 W.
Explanation:
Given;
weight lifted, W = 700 N
height the weight is lifted, h = 2.5 m
time taken to lift the weight, t = 0.60 s
The output power the weightlifter is calculated as;
Power = Energy applied / time taken
Energy applied = weight lifted x height the weight is lifted
Energy applied = 700 x 2.5
Energy applied = 1750 J
Power = 1750 / 0.6
Power = 2916.67 J/s = 2916.67 W.
Therefore, the output power the weightlifter is 2916.67 W.
Answer:
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Explanation:
The probability of getting 4 is 1
Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Missing part of the question: determine the magnitude of Porsha's acceleration.
Given the data in the question;
- Mass of Porsha;

- Mass of Zorn;

- Force of Porsha push;

Magnitude of Porsha's acceleration; 
To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:

Where m is the mass of the object and a is the acceleration.
We substitute the mass of Porsha and the force he used into the equation
Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².
Learn more: brainly.com/question/25125444