Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure, = 2.95 atm
Atmospheric pressure, = 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;
where;
P₁ =
P₂ =
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;
where;
is the density of seawater = 1030 kg/m³
Therefore, the water is flowing at the rate of 28.04 m/s.
Q1=Q2
m1c1(t-t1)=m2c2(t2-t)
67.9kg * c1* (38.7°C-37.1°C)=50.2kg * 4186 J/kg°C * (40.5°C-38.7°C)
67.9kg* c1 * 1.6°C = 50.2kg * 4186 J/kg°C * 1.8°C
108.64 kg°C * c1 = 378246.96 J
c1 = 378246.96J /108.64kg°C
c1=3481.65 J/kg°C
Answer:
I = 69.3 μA
Explanation:
Current through the straight wire, I = 3.45 A
Number of turns, N = 5 turns
Diameter of the coil, D = 1.25 cm
Resistance of the coil,
Distance of the wire from the center of the coil, d = 20 cm = 0.2 m
The magnetic field, B₁, when the wire is at a distance, d, from the center of the coil.
Magnetic field B₂ when the wire is at a distance, 2d from the center of the coil
Change in the magnetic field, ΔB = B₂ - B₁ = 0.00001725 - 0.0000345
ΔB = -0.000001725
Induced current,
E = -N (Δ∅)/Δt
Δ∅ = A ΔB
Area, A = πr²
diameter, d = 0.0125 m
Radius, r = 0.00625 m
A = π* 0.00625²
A = 0.0001227 m²
Δ∅ = -0.000001725 * 0.0001227
Δ∅ = -211.6575 * 10⁻¹²
E = -N (Δ∅)/Δt
Resistance, R = 3.25 μ ohms = 3.25 * 10⁻⁶ ohms
I = E/R
I = 0.0000693 A
I = 69 .3 * 10⁻⁶A
I = 69.3 μA
Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q