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:
Distance: 4.6km Displacement= -0.2km
Explanation:
Total distance: 1.5+2.4+0.7= 4.6 km
Displacement: 1.5-2.4+0.7= -0.2km
The displacement may also be 0.2km, it just depends on if it wants it negative or not.
Answer: E= KQ/r^2
Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.
The magnitude of an electric field E, at a point is given by Coulombs law as
E/ F/q
Where F= Coulombs force exertedon the charge and q= electric charge
E= F/q=(KQq)/r^2q
E=KQ/r^2
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s
Let h₁ = the height at time t, for t ≤ 3.98 s
Let h₂ = the height at time t > 3.98 s
Motion for t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s
Motion for t > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m
The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m
Answer: 931.4 m (nearest tenth)
Answer:
a) 14.2 atm
b) 4.46 atm
c) 1.06 atm
Explanation:
For an ideal gas,
PV = nRT
P = pressure of the gas
V = volume occupied by the gas
n = number of moles of the gas
R = molar gas constant = 0.08206 L.atm/mol.K
T = temperature of the gas in Kelvin
a) For HF,
P =?, V = 2.5L, n = 1.35 moles, T = 320K
P = 1.35 × 0.08206 × 320/2.5
P = 14.2 atm
b) For NO₂
P =?, V = 4.75L, n = 0.86 moles, T = 300K
P = 0.86 × 0.08206 × 300/4.75
P = 4.46 atm
c) For CO₂
P =?, V = 5.5 × 10⁴ mL = 55L, n = 2.15 moles, T = 57°C = 330K
P = 2.15 × 0.08206 × 330/55
P = 1.06 atm
