Parallel-plate capacitor has there fore formula is
<span>C=(<span>ϵ0</span>A)/d
putting values</span>C=(8.85*10^-12*pi*.05^2)/.00063
=1.1*10^-10F
then Q=CV=1.1*10^-10*1000=1.1*10^-7C
as
<span>η=Q/A</span><span>therefore
(1.1*10^-7)/(pi*.05^2)
=1.4*10^-5C/m^ our answer
hope this helps</span>
Answer:
demand (an amount) as a price from someone for a service rendered or goods supplied.
i would say that the child with more linear speed is the cild that is 3 meters away from the center of the merry go round. because the child that is 0.5 meters from the center of the merry go round is less linear because the steering of the merry go round is started from the outer part of the merry go round so it would make more sense that the child that is 3 meters from the center of the merry go round would be more linear in speed.
hope this helps!
Answer:
you start sweating and it would get hot
Answer:
the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
Explanation:
Given the data in the question;
Kinetic energy of each proton that makes up the beam = 3.25 × 10⁻¹⁵ J
Mass of proton = 1.673 × 10⁻²⁷ kg
Charge of proton = 1.602 × 10⁻¹⁹ C
distance d = 2 m
we know that
Kinetic Energy = Charge of proton × Potential difference ΔV
so
Potential difference ΔV = Kinetic Energy / Charge of proton
we substitute
Potential difference ΔV = ( 3.25 × 10⁻¹⁵ ) / ( 1.602 × 10⁻¹⁹ )
Potential difference ΔV = 20287.14 V
Now, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m will be;
E = Potential difference ΔV / distance d
we substitute
E = 20287.14 V / 2 m
E = 10143.57 V/m or 1.01 × 10⁴ V/m
Therefore, the magnitude of a uniform electric field that will stop these protons in a distance of 2 m is 10143.57 V/m or 1.01 × 10⁴ V/m
