Answer:
526.57 Pa
Explanation:
P ( pressure at the bottom of the container) = 1.049 × 10^5 pa
Using the formula of pressure in an open liquid
Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²
Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa
P( atmospheric pressure) = 1.013 × 10^5 Pa
Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g
Subtract each of the pressure from the absolute pressure at the bottom
P(bottom) - atmospheric pressure
(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa
subtract pressure due to water from the remainder
3600 - 2050.29 = 1549.71 Pa
1549.71 = ρ(density of the liquid) × h (depth of the liquid) × g
ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa
The answer is c. +2.0 µC
To calculate this, we will use Coulomb's Law:
F = k*Q1*Q2/r²
where F is force, k is constant, Q is a charge, r is a distance between charges.
k = 9.0 × 10⁹ N*m/C²
It is given:
F = 7.2 N
d = 0.1 m = 10⁻¹ m
Q1 = -4.0 µC = 4 * 1.0 × 10⁻⁶ = 4.0 × 10⁻⁶
Q2 = ?
Thus, let's replace this in the formula for the force:
7.2 = 9.0 × 10⁹ * 4.0 × 10⁻⁶ * Q2/(10⁻¹)²
7.2 = 9 * 4 * 10⁹⁻⁶ * Q2/10⁻¹°²
7.2 = 36 × 10³ * Q2 / 10⁻²
Multiply both sides of the equation by 10⁻²:
7.2 × 10⁻² = 36 × 10³ * Q2
⇒ Q2 = 7.2 × 10⁻² / 36 × 10³ = 7.2/36 × 10⁻²⁻³ = 0.2 × 10⁻⁵ = 2 × 10⁻⁶
Since µC = 1.0 × 10^-6:
Q2 = 2 * 1.0 × 10^-6 = 2 µC
Answer:
to prevent the derailing of the train due to thermal expansion.
Answer:
0 < r < r_exterior B_total = 
r > r_exterior B_total = 0
Explanation:
The magnetic field created by the wire can be found using Ampere's law
∫ B. ds = μ₀ I
bold indicates vectors and the current is inside the selected path
outside the inner cable
B₁ (2π r) = μ₀ I
B₁ =
the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.
For the outer shell
for the case r> r_exterior
B₂ = \frac{\mu_o I}{2\pi r}
This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction
for the case r <r_exterior
in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero
B₂ = 0
Now we can find the field created by each part
0 < r < r_exterior
B_total = B₁
B_total =
r > r_exterior
B_total = B₁ -B₂
B_total = 0
Conception is the correct answer
