Answer:
9.81N
Explanation:
the force of attraction is given by
F=<u>GmM</u><u>/</u><u>R²</u><u> </u>
where m is mass of the body
M is mass of the earth
R is radius of the earth
G is the universal gravitational constant(6.67x10-¹¹)
hence we substitute the values in the formula.
<em> </em><em>you</em><em> </em><em>can</em><em> </em><em>ask</em><em> </em><em>questions</em>
Science is characterized by empirical observations, testable questions, formation of hypotheses and experiments that result in stable and replicable results, logic reasoning and theoretical constructs
Answer:
(a) Charge density σ=6.6375×10²nC/m²
(b) Total charge Q=1.47×10²nC
Explanation:
Given Data
A=47.0 cm =0.47 m
Electric field E=75.0 kN/C
To find
(a) Charge density σ
(b)Total Charge Q
Solution
For (a) charge density σ
From Gauss Law we know that
Φ=Q/ε₀.......eq(i)
Where
Φ is electric flux
Q is charge
ε₀ is permittivity of space
And from the definition of flux
Φ = EA
The flux is electric field passing perpendicularly through the surface
Put the this Φ in equation(i)
EA
=Q/ε₀
where Q(charge)=σA
EA=(σA)/ε₀
E=σ/ε₀
σ=ε₀E

σ=6.6375×10²nC/m²
For (b) total charge Q
Q=σA

Answer:
E = 20.03 J
Explanation:
Given that,
The amount of charge that passes through the filament of a certain lightbulb in 2.00 s is 1.67 C,
Voltage, V = 12 V
We need to find the energy delivered to the lightbulb filament during 2.00 s.
The energy delivered is given by :
. ....(1)
As,

As per Ohm's law, V = IR

Using formula (1).

So, the energy delivered to the lightbulb filament is 20.03 J.