The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
<h3>What do you mean by electric potential? </h3>
The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.
V = kq/r
Where V represents electric potential, K is coulomb constant, q is Charge and r is distance between any two around charge to the point charge.
Electric potential at O due to four charges is given by,
V = 4KQ/ r
where, r = √2a/2 = a/√2
V = 4k × Q√2/a
V = √2Q/πÈa
The total electric potential at the center of the square due to the four charges is V = √2Q/πÈa.
To learn more about electric potential refer to:
brainly.com/question/12645463
#SPJ4
Answer:
A book on its side exerts a greater force.
Explanation:
Pressure = Force / Area
Assuming that 1kg = 10N
2kg = 20N
Area of book lying flat = 0.3m × 0.2m
= 0.6m²
Pressure of book lying flat = 20N / 0.6m²
= 30Pa (1 s.f.)
Area of book on its side = 0.2m × 0.05m
= 0.01m²
Pressure of book on its side = 20N / 0.01m²
= 2000Pa (1 s.f.)
Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.
Answer:
D) 735 J(oules)
Explanation:
Work is defined as force * distance
Force is defined as mass * acceleration
Given a mass of 15 kg and a gravitational acceleration of 9.8 m/s² since the box is being lifted up, the force being applied to the box is 15 kg * 9.8 m/s² = 147 N
Since the distance is 5 meters, the work done is 147 N * 5 m = 735 N/m = 735 J, making D the correct answer.
Answer:
Explanation:
E = σ/ε = (F/A) / (ΔL/L)
E = (mg/(πd²/4) / (ΔL/L)
E = (4mg/(πd²) / (ΔL/L)
E = 4Lmg/(πd²ΔL)
E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)
E = 1.35 x 10⁹ Pa or 1.35 GPa
According to the article "Nuclear shapes" by Renee Lucas the nucleus's shape is mainly modified by vibrational and rotational features happening within the cell. According to the article if i read correctly "near closed shells spherical shapes prevail, while between closed shells the large number of valence nucleons in orbit with large particle angular momentum leads to nuclei with large deformations leading them to not only maintain its shape but also alloying it to work.