Answer:
σ₁ = 
C/m²
σ₂ = 
C/m²
Explanation:
The given data :-
i) The radius of smaller sphere ( r ) = 5 cm.
ii) The radius of larger sphere ( R ) = 12 cm.
iii) The electric field at of larger sphere ( E₁ ) = 358 kV/m. = 358 * 1000 v/m
Q₁ = 572.8 
C
Since the field inside a conductor is zero, therefore electric potential ( V ) is constant.
V = constant
∴
= 
C
Surface charge density ( σ₁ ) for large sphere.
Area ( A₁ ) = 4 * π * R² = 4 * 3.14 * 0.12 = 0.180864 m².
σ₁ = 
= 
= C/m².
Surface charge density ( σ₂ ) for smaller sphere.
Area ( A₂ ) = 4 * π * r² = 4 * 3.14 * 0.05² =0.0314 m².
σ₂ =
= 
= C/m²
Answer
given,
net charge = +2.00 μC
we know,
1 coulomb charge = 6.28 x 10¹⁸electrons
1 micro coulomb charge = 6.28 x 10¹⁸ x 10⁻⁶ electron
= 6.28 x 10¹² electrons
2.00 μC = 2 x 6.28 x 10¹² electrons
= 1.256 x 10¹³ electrons
since net charge is positive.
The number of protons should be 1.256 x 10¹³ more than electrons.
hence, +2.00 μC have 1.256 x 10¹³ more protons than electrons.
Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa
Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar
1000 kcal because you only get 10% of the energy of the thing you eat
