The density of silver is ρ = 10500 kg/m³ approximately.
Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet
Let A be the area.
Then, by definition,
m = (t*A)*ρ
Therefore
A = m/(t*ρ)
= (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
= 539.7 m²
Answer: 539.7 m²
The sphere’s Electric potential energy is 1.6*
J
Given,
q=6. 5 µc, V=240 v,
We know that sphere’s Electric potential energy(E) = qV=6.5*
=1.6*
J
<h3>Electric potential energy</h3>
The configuration of a certain set of point charges within a given system is connected with the potential energy (measured in joules) known as electric potential energy, which is a product of conservative Coulomb forces. Two crucial factors—its inherent electric charge and its position in relation to other electrically charged objects—can determine whether an object has electric potential energy.
In systems with time-varying electric fields, the potential energy is referred to as "electric potential energy," but in systems with time-invariant electric fields, the potential energy is referred to as "electrostatic potential energy."
A tiny sphere carrying a charge of 6. 5 µc sits in an electric field, at a point where the electric potential is 240 v. what is the sphere’s potential energy?
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Yessss how many you have very interesting
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm