A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² -
r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄
B) E = kQ/d²
since the distribution is symmetric spherically
C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 -
r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.
D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a
min).
<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>
- A 16.0 kg canoe moving to the left at 12.5 m/s makes an elastic head on collision with a 14.0 kg raft moving to the right at 16.0 m/s.
- After the collision the raft moves to the left at 14.4 m/s assuming water simulates a frictionless surface.
- Mass of the canoe (m1) = 16 Kg
- Mass of the raft (m2) = 14 Kg
- Initial velocity of the canoe (u1) = 12.5 m/s
- Initial velocity of the raft (u1) = - 16 m/s [Here, the raft's velocity is negative, because the objects are moving in the opposite direction]
- Total momentum of the system = m1u1 + m2u2 = [(16 × 12.5) + (14 × -16)] Kg m/s = (200 - 224) Kg m/s = -24 Kg m/s
- Final velocity of the raft (v2) = 14.4 m/s
- Let the final velocity of the canoe be v1.
- Total momentum of the system after the impact = m1v1 + m2v2 = [(16 × v1) + (14 × 14.4)] Kg m/s = 16v1 Kg + 201.6 Kg m/s
- According to the law of conservation of momentum, Total momentum of the system before the impact = Total momentum of the system after the impact
- or, -24 Kg m/s = 16v1 Kg + 201.6 Kg m/s
- or, -24 Kg m/s - 201.6 Kg m/s = 16v1 Kg
- or, -225.6 Kg m/s = 16v1 Kg
- or, v1 = -225.6 Kg m/s ÷ 16 Kg
- or, v1 = -14.1 m/s
<u>Answer:</u>
<u>T</u><u>he final velocity of the </u><u>canoe </u><u>is </u><u>-</u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>or </u><u>1</u><u>4</u><u>.</u><u>1</u><u> </u><u>m/</u><u>s </u><u>to </u><u>the </u><u>right.</u>
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
a) m₁ = 1.41 kg
, b) m₂ = 2.65 kg
Explanation:
For this exercise we will use Newton's second law
Block 1
T - W₁ = m₁ a
Block 2
W₂ - T = m₂ a
We have selected the positive block 1 rising and block two lowering, as the pulley has no friction does not affect the movement
Let's use kinematics to look for acceleration
y = v₀ t + ½ a t²
As part of the rest the initial speed is zero
a = 2 y / t²
a = 2 6.00 / 2²
a = 3 m / s²
Let's replace in the equation of block 1
a) T = m₁ g + m₁ a
m₁ = T / (g + a)
m₁ = 18.0 / (9.8 + 3)
m₁ = 1.41 kg
b) we substitute in the equation of block 2
W₂ - T = m₂ a
m₂ g - m₂ a = T
m₂ = T / (g-a)
m₂ = 18.0 / (9.8 -3)
m₂ = 2.65 kg
A modern statement of Charles's law is: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion. ... The equation shows that, as absolute temperature increases, the volume of the gas also increases in proportion.
i hope it will help