Answer:
A. Aluminium
Explanation:
Buoyant force is simple terms refers to the weight of water displaced when an object is immersed in it. The higher the volume of an object, the higher the buoyant force.
From the problem, we have two metals of equal mass, aluminium and lead, of which lead is denser.
density =
Volume =
This means that the volume of each metal depends on the density, and for the fact that lead is denser , it will have a lower volume than aluminium which means Aluminium will have a greater buoyant force.
No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.
Incomplete question as the charge density is missing so I assume charge density of 3.90×10^−12 C/m².The complete one is here.
An electron is released from rest at a distance of 0 m from a large insulating sheet of charge that has uniform surface charge density 3.90×10^−12 C/m² . How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00×10−2 m from the sheet?
Answer:
Work=1.06×10⁻²¹J
Explanation:
Given Data
Permittivity of free space ε₀=8.85×10⁻¹²c²/N.m²
Charge density σ=3.90×10⁻¹² C/m²
The electron moves a distance d=3.00×10⁻²m
Electron charge e=-1.6×10⁻¹⁹C
To find
Work done
Solution
The electric field due is sheet is given as
E=σ/2ε₀
Now we need to find force on electron
Now for Work done on the electron
Answer:
Since the wire is not splitting at any point in the circuit,
the resistors are in series
Hence, Equivalent resistance = 10 + 20 + 30
Equivalent Resistance = 60 Ω
Answer:
10573375000
Explanation:
k = Coulomb constant = 
r = Distance = 
E = Electric field = 1150 N/C
Electric field is given by
Number of electrons is given by
Number of excess electrons is 10573375000
r = 0.115+0.15 = 0.265 m
The electric field is
