Which of these parts reduces the amount of pollutants in the engine exhaust?
A. Transmission
B. Catalytic converter
C. Tailpipe
D. Muffler

Answer:

car a is moving faster than the car b

Answer:

B: Car A and car B are moving in opposite directions.

C: Car A is moving faster than car B.

E: Car A and car B are moving toward each other until they cross over.

Explanation:

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Answer:eyesight

Explanation:

Solution :

import [tex]$\text{java}.$[/tex]util.*;

public [tex]$class$[/tex] currency{

  public static [tex]$\text{void}$[/tex] main(String[tex]$[]$[/tex] args) {

      Scanner input [tex]$=$[/tex] new Scanner(System[tex]$\text{.in}$[/tex]);

      System[tex]$\text{.out.}$[/tex]print("Enter [tex]$\text{number of}$[/tex] quarters:");

      int quarters = input.nextInt();

      System.out.print("Enter number of dimes:");

      int [tex]$\text{dimes =}$[/tex] input.nextInt();

      System[tex]$\text{.out.}$[/tex]print("Enter number of nickels:");

      int nickels = input.nextInt();

      System[tex]$\text{.out.}$[/tex]print("Enter number of pennies:");

      int [tex]$\text{pennies = }$[/tex] input.nextInt();

      // computing dollors

      double dollars = (double) ((quarters*0.25)+(dimes*0.10)+(nickels*0.05)+(pennies*0.01));

      System[tex]$\text{.out.}$[/tex]format("You have : $%.2f",dollars);

  }

}

Answer: its c

Explanation:

Answer:

I am in Eight Grade

Explanation:

Answer:

print(100/42)

Explanation:

This is the operand for division in python and some other languages.

count ← 0

REPEAT 4 TIMES

{

count count + 1

REPEAT count TIMES

{

MOVE_FOWARD ()

}

ROTATE_LEFT ()

}

The message is broken into bundles that are communicated in a predefined request. Every bundle should be gotten in the request it was sent for the message to be accurately reassembled by the beneficiary's gadget.

A user installs a program on their computer that records and transmits all user input unintentionally.

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Answer: 50

Explanation:

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature [tex]T_1 = -16^0\ C[/tex]

Quality [tex]x_1 = 0.2[/tex]

Outlet:

Temperature [tex]T_2 = -16^0 C[/tex]

Quality  [tex]x_2 = 1[/tex]

The following data were obtained at saturation properties of R134a at the temperature of -16° C

[tex]v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\ v_g = 0.1247 \ m^3 /kg[/tex]

[tex]v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\ v_1 = 0.0255 \ m^3/kg \\ \\ \\ v_2 = v_g = 0.1247 \ m^3/kg[/tex]

[tex]m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6 \\ \\ \mathbf{m = 0.0534 \ kg/sec}[/tex]

[tex]\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2 \\ \\ \rho_1v_1 = \rho_2v_2 \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}[/tex]

Answer: True

Explanation:

Structural engineering is simply referred to as a branch of civil engineering, and they help in the designing of structures like buildings, dams, tunnels, bridges, tunnels, etc. Structural engineers typically work as consultants.

Wireless sensor networks are used by structural engineers to monitor the condition of dams and bridges. Wireless sensor network refers are dedicated sensors that are used in monitoring the physical conditions of the bridges and to know if they're still in good conditions.

Answer: c

Explanation:

Answer:

Hermetic compressors are ideal for small refrigeration systems, where continuous maintenance cannot be ensured.

Answer:

"150000 N/m²" is the right approach.

Explanation:

According to the question, the pressure on the two spheres 1 and 2 is same.

Sphere 1 and 2:

Then,

⇒  [tex]P_1=P_2[/tex]

⇒  [tex]\frac{\Delta V_1}{V_1}=\frac{\Delta V_2}{V_2}[/tex]

and the bulk modulus be,

⇒  [tex]B_1=B_2[/tex]

Sphere 3:

⇒  [tex]\frac{\Delta V_3}{V_3} =\frac{\frac{\Delta V_1}{V_1} }{\frac{\Delta V_2}{V_2} } =1[/tex]

then,

⇒  [tex]P_3=B\times \frac{\Delta V_3}{V_3}[/tex]

⇒       [tex]=B\times 1[/tex]

⇒       [tex]=150000\times 1[/tex]

⇒       [tex]=150000 \ N/m^2[/tex]

Answer:

I) Voltage - is the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as illuminating a light. In brief, voltage = pressure, and it is measured in volts (V).

II) Current - is the movement of electrons through a wire. Electric current is measured in amperes (amps) and refers to the number of charges that move through the wire per second. If we want current to flow directly from one point to another, we should use a wire that has as little resistance as possible.

III) Electrical Power - is the rate, per unit time, at which electrical energy is transferred by an electric circuit. The SI unit of power is the watt, one joule per second. Electric power is usually produced by electric generators, but can also be supplied by sources such as electric batteries.

IV) Electrical Energy - is a form of energy resulting from the flow of electric charge. Energy is the ability to do work or apply force to move an object. In the case of electrical energy, the force is electrical attraction or repulsion between charged particles.

Explanation:

I hope ot helps to you a lot! Correct me if I'm wrong.

Answer:

Sin(angle)=0.1538

Angle=arcsine(0.1538)

Angle is 8.85 degrees

Explanation:

Firstly draw a diagram showing a triangle with the required angle, a hypotenuse of 130 ft and an opposite side of 20 ft.

The sine of the angle required uses the formula Opposite/Hypotenuse

(i am not sure sure)

Electronic rack-and-pinion systems use an electric motor to pump the hydraulic fluid through the hoses. This statement is true.

The rack and pinion gearset is housed inside a metal tube, with each end of the rack protruding out and being attached to an axial rod. This is how it operates. When the steering wheel is twisted, the rack is moved by the pinion gear, which is mounted to the steering shaft. Based on the torque being given to the steering wheel by the driver, the steering wheel position, and the vehicle's speed, the EPS electronic control unit (ECU) determines the amount of aiding power required. With an applied force, the EPS motor rotates a steering gear, lowering the driver's necessary torque.

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Answer: barries

Explanation:

Answer:

(T²/D³)sys1 = (T²/D³)sys2

sys1 = earth-moon

sys2 = earth-sat

(27.33day)²/(3.8e8m)³ = (1day)²/D³

D = cbrt(7.3e22m³) = 4.2e7 m

Explanation:

exchange in capacity whilst a satellite tv for pc differences altitude How plenty paintings could be executed to flow the satellite tv for pc into yet another around orbit it is bigger above the outdoors of the Earth? satellite tv for pc exchange in capacity with top Assuming the satellite tv for pc is to be boosted to a clean top r? Gravitational capacity capacity (to center of earth) new orbit(2) has a greater robust PE than old one(a million), so exchange is helpful PE = G m?m?/r earth GM = 3.98e14 ?PE = (GM)(m)(a million/r? – a million/r?) KE additionally differences. Get speed at each and every top. New orbit(2) has decrease speed, so exchange is damaging v = ?(GM/R) V? = ?(GM/r?) V? = ?(GM/r?) ?KE = –½m(V?² – V?²) ?KE = –½mGM(a million/r? – a million/r?) including the two ?E = (GM)(m)(a million/r? – a million/r?)– ½mGM(a million/r? – a million/r?) ?E = ½mGM(a million/r? – a million/r?) ?E = ½(3.98e14)(7500) [a million/(0.5e7) –a million/(3.3e7) ] ?E = ½(3.98e14)(7500)(1e-7) [a million/(0.5) –a million/(3.3) ] ?E = ½(3.98e7)(7500) [2 – 0.303 ] ?E = ½(3.98e7)(7500)(a million.70) ?E = 2.04e11 Joules edit, corrected .

The radius of a circular orbit will be "[tex]\frac{V}{2 \pi} \ km[/tex]".

According to the question,

Time = 1 day

Distance = [tex]2 \pi r[/tex]

Let,

As we know,

→ [tex]Distance = Velocity\times time[/tex]

By substituting the values, we get

→          [tex]2 \pi r = V\times 1[/tex]

→              [tex]r = \frac{V}{2 \pi} \ km[/tex]

Thus the above is the right answer.

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This question is incomplete, The missing image is uploaded along this answer below;

Answer:

the maximum stress in the cylinder is 3.23 ksi

Explanation:

Given the data in the question and the diagram below;

First we determine the initial Kinetic Energy;

T = [tex]\frac{1}{2}[/tex]mv²

we substitute

⇒ T = [tex]\frac{1}{2}[/tex] × (550/32.2) × (2)²

T = 34.16149 lb.ft

T =  ( 34.16149 × 12 ) lb.in  

T = 409.93788 lb.in

Now, the volume will be;

V = [tex]\frac{\pi }{4}[/tex]d²L

from the diagram; d = 0.5 ft and L = 1.5 ft

so we substitute

V =  [tex]\frac{\pi }{4}[/tex] × ( 0.5 × 12 in )² × ( 1.5 × 12 in )

V = 508.938 in³

So by conservation of energy;

Initial energy per unit volume = Strain energy per volume

⇒ T/V = σ²/2E

from the image; E = 6.48(10⁶) kip

so we substitute

⇒ 409.93788 / 508.938 = σ²/2[6.48(10⁶)]

508.938σ² =  5,312,794,924.8

σ² = 10,438,982.5967  

σ = √10,438,982.5967

σ = 3230.9414  

σ = 3.2309 ksi  ≈ 3.23 ksi    { three significant figures }

Therefore, the maximum stress in the cylinder is 3.23 ksi

Answer: its c

Explanation:

Answer:

1279

Explanation:

We have the mean u = 1497

Standard deviation sd = 322

We find the x distribution using 25%

P(Z<z) = 0.25

Z = -0.675

From here we use the formula for z score

X = z(sd) + u

X = -0.675*322 + 1497

X = -217.35 + 1497

X = 1279.6

Which is approximately 1279

So we conclude that the maximum sat scores in year 2014that meets with the requirements of this course is 1279

Answer:

1831

Explanation:

Answer:

Explanation:

From the information given:

original diameter [tex]d_o[/tex] = 10 mm

final diameter [tex]d_f =[/tex] 7.5 mm

Cold work tensile strength of brass = 380 MPa

Recall that;

[tex]\text {The percentage CW }= \dfrac{\pi (\dfrac{d_o}{2})^2 - \pi(\dfrac{d_f}{2})^2 }{\pi(\dfrac{d_o}{2})^2} \times 100[/tex]

[tex]\implies \dfrac{\pi (\dfrac{10}{2})^2 - \pi(\dfrac{7.5}{2})^2 }{\pi(\dfrac{10}{2})^2} \times 100[/tex]

[tex]\implies43.87\% \ CW[/tex]

→ At 43.87% CW, Brass has a tensile strength of around 550 MPa, which is greater than 380 MPa.

→ At 43.87% CW, the ductility is less than 5% EL, As a result, the conditions aren't met.

To achieve 15% EL, 28% CW is allowed at most

i.e

The lower bound cold work = 15%

The upper cold work = 28%

The average = [tex]\dfrac{15+28}{2}[/tex] = 21.5 CW

Now, after the first drawing, let the final diameter be [tex]d_o^'[/tex]; Then:

[tex]4.5\% \ CW = \dfrac{\pi (\dfrac{d_o^'}{2})^2 - (\dfrac{7.5}{2})^2}{\pi (\dfrac{d_o^'}{2})^2}\times 100[/tex]

By solving:

[tex]d_o^'} = 8.46 mm[/tex]

To meet all of the criteria raised by the question, we must first draw a wire with a diameter of 8.46 mm and then 21.5 percent CW on it.

Answer:  is b

Explanation:

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / [tex]0.02^{0.22[/tex]

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

[tex]l_i[/tex] = [tex]l_0e^{ET[/tex]

given that [tex]l_0[/tex] = 480 mm

we substitute

[tex]l_i[/tex] =[tex]480mm[/tex] × [tex]e^{0.04481[/tex]

[tex]l_i[/tex] =  501.998 mm

Now we find the elongation;

Elongation = [tex]l_i - l_0[/tex]

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

Answer:

The law of parallelogram of forces states that if two vectors acting on a particle at the same time be represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point their resultant vector is represented in magnitude and direction by the diagonal of the parallelogram drawn from .

Answer:

The strain of the steel rod due to the applied force is 41.93

Explanation:

Modulus of elasticity is equal to stress divided by strain.

And stress is equal to force divided by area

Surface area of cylindrical rod

[tex]2\pi r (r+h)[/tex]

Substituting the given values we get -

[tex]2 *3.14 * \frac{1.5}{1000} * 45 (45 + \frac{1.5}{1000}) = 19.07[/tex]

[tex]2 * 10 ^5 = \frac{250}{19.07 * S=(\frac{\Delta L}{L} )}[/tex]

Hence, strain is equal to

Strain = 41.93

Answer: its c

Explanation: