The electric field of a very large (essentially infinitely large) plane of charge is given by:
E = σ/(2ε₀)
E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.
To determine σ:
σ = Q/A
Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:
σ = Q/d²
Make this substitution in the equation for E:
E = Q/(2ε₀d²)
We see that E is inversely proportional to the square of d:
E ∝ 1/d²
The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

Answer:
μ = 0.33
Equal to 3.2 m/s²
Explanation:
Draw a free body diagram of the block. There are three forces:
Normal force N pushing up.
Weight force mg pulling down.
Friction force Nμ pushing opposite the direction of motion.
Sum of forces in the y direction.
∑F = ma
N − mg = 0
N = mg
Sum of forces in the x direction.
∑F = ma
Nμ = ma
Substitute.
mgμ = ma
μ = a/g
μ = (3.2 m/s²) / (9.8 m/s²)
μ = 0.33
As found earlier, the acceleration is a = gμ. Since g and μ are constant, a is also constant, so it does not change with velocity.
One way is speed=distance divided by time
Answer with Explanation:
We are given that
Initial velocity,u=4.5 m/s
Time=t =0.5 s
Final velocity=v=0m/s
We have to find the deceleration and estimate the force exerted by wall on you.
We know that
Acceleration=
Using the formula
Acceleration=
deceleration=a=
We know that
Force =ma
Using the formula and suppose mass of my body=m=40 kg
The force exerted by wall on you
Force=
Answer:
If you are using a machine that has a mechanical advantage of 2.5,you will have increased your EFFORT DISTANCE by 2.5 times. This allows you to REDUCE the EFFORT FORCE needed by 2.5 times.