All categories

2 years ago

Hii I need help can someone help me

Engineering

1 answer:

zavuch27 [327]2 years ago

3 0

Answer:

I'm not 100 percent sure, but I'm pretty sure that it is half of an ellipse.

Explanation:

For one, if you look at a picture of a hyperbola, you can see that it looks like nothing and is nothing compared to the parabola. However, looking at an ellipse, you can clearly see the resemblance between the two.

Once again, I am not 100 percent certain with this answer, so if you have anything to say about it please let me know.

Have a nice day!

You might be interested in

At an impaired driver checkpoint, the time required to conduct the impairment test varies (according to an exponential distribut

professor190 [17]

Answer:

Option (d) 2 min/veh

Explanation:

Data provided in the question:

Average time required = 60 seconds

Therefore,

The maximum capacity that can be accommodated on the system, μ = 60 veh/hr

Average Arrival rate, λ = 30 vehicles per hour

Now,

The average time spent by the vehicle is given as

⇒ $\frac{1}{\mu(1-\frac{\lambda}{\mu})}$

thus,

on substituting the respective values, we get

Average time spent by the vehicle = $\frac{1}{60(1-\frac{30}{60})}$

Average time spent by the vehicle = $\frac{1}{60(1-0.5)}$

Average time spent by the vehicle = $\frac{1}{60(0.5)}$

Average time spent by the vehicle = $\frac{1}{30}$ hr/veh

Average time spent by the vehicle = $\frac{1}{30}\times60$ min/veh

[ 1 hour = 60 minutes]

thus,

Average time spent by the vehicle = 2 min/veh

Hence,

Option (d) 2 min/veh

7 0

3 years ago

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator

MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit $i_m=385mA=385\times 10^{-3}A=0.385A$

Inductance of the inductor $L=400mH=400\times 10^{-3}h=0.4H$

Capacitance $C=4.43\mu F=4.43\times 10^{-3}F$

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be $x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm$

Capacitive reactance will be equal to $X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm$

Impedance of the circuit will be $Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm$

So maximum voltage will be $\Delta V_{max}=0.385\times 865.44=333.194volt$

(B) Phase difference will be given as $\Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70$

So current will be leading by an angle 54.70

5 0

2 years ago

What is engine knock? What cause the engine knock problem?

antiseptic1488 [7]

Answer:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

Explanation:

When the uneven burning of the fuel takes place due to the incorrect air/fuel mixture inside the engine cylinder, a knocking sound is observed. This is called as the engine knocking.

The engine knock problem can be caused due to the following reason

a) When the octane rating of the fuel used is low.

b) The deposition of the carbon around the cylinder walls takes place.

c) The spark plug used in the vehicle is not correct.

3 0

2 years ago

You are using a Jupyter Notebook to explore data in a DataFrame named productDF. You want to write some inline SQL by using the

defon

You need to explain it more simple as everyone is clueless

7 0

2 years ago

Determine the required dimensions of a column with a square cross section to carry an axial compressive load of 6500 lb if its l

ycow [4]

Answer: 0.95 inches

Explanation:

A direct load on a column is considered or referred to as an axial compressive load. A direct concentric load is considered axial. If the load is off center it is termed eccentric and is no longer axially applied.

The length= 64 inches

Ends are fixed Le= 64/2 = 32 inches

Factor Of Safety (FOS) = 3. 0

E= 10.6× 10^6 ps

σy= 4000ps

The square cross-section= ia^4/12

PE= π^2EI/Le^2

6500= 3.142^2 × 10^6 × a^4/12×32^2

a^4= 0.81 => a=0.81 inches => a=0.95 inches

Given σy= 4000ps

σallowable= σy/3= 40000/3= 13333. 33psi

Load acting= 6500

Area= a^2= 0.95 ×0.95= 0.9025

σactual=6500/0.9025

σ actual < σallowable

The dimension a= 0.95 inches

3 0

2 years ago

Read 2 more answers