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Anvisha [2.4K]
3 years ago
11

While having a discussion, Technician A says that you should never install undersized tires on a vehicle. The vehicle will be lo

wer, and the speedometer will no longer be accurate. Technician B says that the increase in engine rpm for a given speed will result in a decrease in fuel economy. Who is correct
Engineering
1 answer:
gulaghasi [49]3 years ago
6 0

Answer:

Both technician A and technician B are correct.

Explanation: Vehicle manufacturers always specify the size of the tires required for a given vehiclefor optimal efficiency,this will ensure that the speedometer is accurate and the level of the vehicle is good enough to ensure the vehicle works efficiently.

It is also a known fact that an increase in a vehicle's rpm(revolution per minute) will eventually lead to increased fuel consumption which means the fuel economy of the vehicle will be reduced making the vehicle less efficient in its fuel consumption.

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What is the output of a system with the transfer function s/(s + 3)^2 and subject to a unit step input at time t = 0?
Dominik [7]

Answer:

0

Explanation:

output =transfer function H(s) ×input U(s)

here H(s)=\frac{s}{(s+3)^2}

U(s)=\frac{1}{s} for unit step function

output =H(s)×U(s)

=\frac{s}{(s+3)^2}×\frac{1}{s}

=\frac{1}{(s+3)^2}

taking inverse laplace of output

output=t×e^{-3t}

at t=0 putting the value of t=0 in output

output =0

3 0
4 years ago
soccer is also called association football" A soccer ball is a sphere, with circumference of 70 centimeters. in developing a new
timama [110]

Answer: Weight on Mars = 0.02593N

Explanation:

Given; Circumference C of Sphere = 70cm = 0.7m,

Specific Gravity S. G. of material = 1.21,

acceleration due to gravity in the Mars gm = 3.7m/s^2

We know that Weight W = mass m × acceleration due to gravity.

Let the Weight in on the Mars be Wm.

Wm = m × gm

Since we are given gm, we need to calculate for m. (Note that mass m is the same everywhere)

But mass = specific gravity × volume

Since we know the specific gravity, let's go ahead to calculate for the volume of the ball.

We know that Volume of a Sphere V = (4/3)πr^3

To get r, we know that C = 2πr

Therefore, r = C/(2π) = 0.7/(2π) = (7/10)/2π = 7/20π (in meters)

V = (4/3)*π×(7/20π)^3 = 343/6000π^2 (in meter^3)

m = 343/6000π^2 × 1.21 = 7.01×10^(-3)kg

Wm = 7.01×10^(-3) × 3.7 = 0.02593N

8 0
3 years ago
Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th
jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

5 0
3 years ago
Air is compressed in a reversible, isothermal, steady- flow process from 15 psia, 100°F to 100 psia. Calculate the work of compr
mixas84 [53]

Answer:

|W|=169.28 KJ/kg

ΔS = -0.544 KJ/Kg.K

Explanation:

Given that

T= 100°F

We know that

1 °F = 255.92 K

100°F = 310 .92 K

P _1= 15 psia

P _1= 100 psia

We know that work for isothermal process  

W=mRT\ln \dfrac{P_1}{P_2}

Lets take mass is 1 kg.

So work per unit mass

W=RT\ln \dfrac{P_1}{P_2}

We know that for air R=0.287KJ/kg.K

W=RT\ln \dfrac{P_1}{P_2}

W=0.287\times 310.92\ln \dfrac{15}{100}

W= - 169.28 KJ/kg

Negative sign indicates compression

|W|=169.28 KJ/kg

We know that change in entropy at constant volume

\Delta S=-R\ln \dfrac{P_2}{P_1}

\Delta S=-0.287\ln \dfrac{100}{15}

ΔS = -0.544 KJ/Kg.K

3 0
3 years ago
Which one? Sam Golbach or Colby Brock
Maru [420]
Sam all the way.....
5 0
3 years ago
Read 2 more answers
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