Answer:
30 miles
Explanation:
<u>Step 1:</u>
Divide -> 45/60= .75 miles/minute
<u>Step 2:</u>
Multiply -> .75 x 40= 30
Answer:
I = 21.13 mA ≈ 21 mA
Explanation:
If
I₁ = 5 mA
L₁ = L₂ = L
V₁ = V₂ = V
ρ₁ = 1.68*10⁻⁸ Ohm-m
ρ₂ = 1.59*10⁻⁸ Ohm-m
D₁ = D
D₂ = 2D
S₁ = 0.25*π*D²
S₂ = 0.25*π*(2*D)² = π*D²
If we apply the equation
R = ρ*L / S
where (using Ohm's Law):
R = V / I
we have
V / I = ρ*L / S
If V and L are the same
V / L = ρ*I / S
then
(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂
If
S₁ = 0.25*π*D² and
S₂ = 0.25*π*(2*D)² = π*D²
we have
ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)
⇒ I₂ = 4*ρ₁*I₁ / ρ₂
⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m
⇒ I₂ = 21.13 mA
Answer:
(a) 0.063 m/s
(b) 1.01 m/s
Explanation:
rate of volume flow, V = 4 x 10^-6 m^3/s
(a) radius, r = 4.5 x 10^-3 m
Let the speed of blood is v.
So, V = A x v
where A be the area of crossection of artery
4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v
v = 0.063 m/s
Thus, the speed of flow of blood is 0.063 m/s .
(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m
Let the speed is v'.
So, V = A' x v'
4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'
v' = 1.01 m/s
Thus, the speed of flow of blood is 1.01 m/s .
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is

a. The angular acceleration is Torque divided by moments of inertia:

b. 5 revolution would be equals to
rad, or 31.4 rad. Since the engine just got started


c. Work done during the first 5 revolution would be torque times angular displacement:

d. The time it takes to spin the first 5 revolutions is

The average power output is work per unit time
or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
or 84 kW
Answer: 7.53 μC
Explanation: In order to explain this problem we have to use the gaussian law so we have:
∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0
Q inside= 0 = q+ Qinner surface=0
Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC