Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
__
<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
__
<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Answer:
a) the metal removal rate is 14.4 in³/min
b) the cutting time is 0.98 min
Explanation:
Given the data from the question
first we find the rpm for the spindle of the drilling tool, using the equation
Ns = 12V/πD
V is the cutting speed(120 fpm) and D is the diameter of the hole( 2 in)
so we substitute
Ns = 12 × 120 / π2
Ns = 1440 / 6.2831
Ns = 229.18 rmp
Now we find the metal removal rate using the equation
MRR = (πD²/4) Fr × Ns
Fr is the feed rate( 0.02 ipr ),
so we substitute
MRR = ((π × 2²)/4) × 0.02 × 229.18
MRR = 14.3998 ≈ 14.4 in³/min
Therefore the metal removal rate is 14.4 in³/min
Next we find the allowance for approach of the tip of the drill
A = D/2
A = 2/2
= 1 in
now find the time required to drill the hole
Tm = (L + A) / (Fr × Ns)
Lis the the depth of the hole( 3.5 in)
so we substitute our values
Tm = (3.5 + 1) / (0.02 × 229.18 )
Tm = 4.5 / 4.5836
Tm = 0.98 min
Therefore the cutting time is 0.98 min
Answer:
(a) E = 0 N/C
(b) E = 0 N/C
(c) E = 7.78 x10^5 N/C
Explanation:
We are given a hollow sphere with following parameters:
Q = total charge on its surface = 23.6 μC = 23.6 x 10^-6 C
R = radius of sphere = 26.1 cm = 0.261 m
Permittivity of free space = ε0 = 8.85419 X 10−12 C²/Nm²
The formula for the electric field intensity is:
E = (1/4πεo)(Q/r²)
where, r = the distance from center of sphere where the intensity is to be found.
(a)
At the center of the sphere r = 0. Also, there is no charge inside the sphere to produce an electric field. Thus the electric field at center is zero.
<u>E = 0 N/C</u>
(b)
Since, the distance R/2 from center lies inside the sphere. Therefore, the intensity at that point will be zero, due to absence of charge inside the sphere (q = 0 C).
<u>E = 0 N/C</u>
(c)
Since, the distance of 52.2 cm is outside the circle. So, now we use the formula to calculate the Electric Field:
E = (1/4πεo)[(23.6 x 10^-6 C)/(0.522m)²]
<u>E = 7.78 x10^5 N/C</u>
Answer: At time 18.33 seconds it will have moved 500 meters.
Explanation:
Since the acceleration of the car is a linear function of time it can be written as a function of time as
Integrating both sides we get
Now since car starts from rest thus at time t = 0 ; v=0 thus c=0
again integrating with respect to time we get
Now let us assume that car starts from origin thus D=0
thus in the first 15 seconds it covers a distance of
Thus the remaining 125 meters will be covered with a constant speed of
in time equalling 
Thus the total time it requires equals 15+3.33 seconds
t=18.33 seconds
What am I going to select?? What are my choices bro????
