1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Natalija [7]
3 years ago
14

How do you change a car tire ?

Engineering
2 answers:
stiks02 [169]3 years ago
8 0
You remove the old one and put a new one.
ladessa [460]3 years ago
8 0

Answer:

Explanation:

Find a Safe Location. As soon as you realize you have a flat tire, do not abruptly brake or turn. ...

Turn on Your Hazard Lights. ...

Apply the Parking Brake. ...

Apply Wheel Wedges. ...

Remove the Hubcap or Wheel Cover. ...

Loosen the lug nuts. ...

Place the Jack Under the Vehicle. ...

Raise the Vehicle With the Jack.

And i got it from google idk how to change a tire and hope this helps you out and mark me as brainiest if u think its a great answer

-sammantha

You might be interested in
A gasoline engine has a piston/cylinder with 0.1 kg air at 4 MPa, 1527◦C after combustion, and this is expanded in a polytropic
Roman55 [17]

Answer:

The expansion work is 71.24 kJ and heat transfer is -16.89 kJ

Explanation:

From ideal gas law,

Initial volume (V1) = nRT/P

n is the number of moles of air in the cylinder = mass/MW = 0.1/29 = 0.00345 kgmol

R is gas constant = 8314.34 J/kgmol.K

T is initial temperature = 1527 °C = 1527+273 = 1800 K

P is initial pressure = 4 MPa = 4×10^6 Pa

V1 = 0.00345×8314.34×1800/(4×10^6) = 0.013 m^3

V2 = 10×V1 = 10×0.013 = 0.13 m^3

The process is a polytropic expansion process

polytropic exponent (n) = 1.5

P2 = P1(V1/V2)^n = 4×10^6(0.013/0.13)^1.5 = 1.26×10^5 Pa

Expansion work = (P1V1 - P2V2) ÷ (n - 1) = (4×10^6 × 0.013 - 1.26×10^5 × 0.13) ÷ (1.5 - 1) = 35620 ÷ 0.5 = 71240 J = 71240/1000 = 71.24 kJ

Heat transfer = change in internal energy + expansion work

change in internal energy (∆U) = Cv(T2 - T1)

T2 = PV/nR = 1.26×10^5 × 0.13/0.00345×8314.34 = 571 K

Cv = 20.785 kJ/kgmol.K

∆U = 20.785(571 - 1800) = -25544.765 kJ/kgmol × 0.00345 kgmol = -88.13 kJ

Heat transfer = -88.13 + 71.24 = -16.89 kJ

5 0
3 years ago
Click this link to view O*NET’s Wages and Employment section for Film and Video Editors.
vova2212 [387]

Answer:

much faster than average

Explanation:

did it on edge (2022-2032)

6 0
2 years ago
Write torsion equation and explain the importance of each components.<br>​
Elanso [62]
The equations are based on the following assumptions

1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.

Nomenclature

T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)

4 0
2 years ago
An 800-kg drag racer accelerates from rest to 390 km/hr in 5.8 s. What is the net impulse applied to the racer in the first 5.8
marissa [1.9K]

Answer:

Impulse =14937.9 N

tangential force =14937.9 N

Explanation:

Given that

Mass of car m= 800 kg

initial velocity u=0

Final velocity v=390 km/hr

Final velocity v=108.3 m/s

So change in linear momentum P= m x v

           P= 800 x 108.3

 P=86640 kg.m/s

We know that impulse force F= P/t

So F= 86640/5.8 N

F=14937.9 N

Impulse force F= 14937.9 N

We know that

v=u + at

108.3 = 0 + a x 5.8

a=18.66\ m/s^2

So tangential force F= m x a

F=18.66 x 800

F=14937.9 N

6 0
3 years ago
Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of
yKpoI14uk [10]

Answer: Hello the question is incomplete below is the missing part

Question:  determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

V2 = 600 ft/s

V1 = 60 ft/s

h1 = 129.0613  ( value gotten from Ideal gas property-air table )

<em>first step : calculate the value of h2 using the equation below </em>

assuming no work is done ( potential energy is ignored )

h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

we will apply interpolation

attached below is the remaining part of the solution

8 0
2 years ago
Other questions:
  • A slight breeze is blowing over the hot tub above and yields a heat transfer coefficient h of 20 W/m2 -K. The air temperature is
    15·1 answer
  • When a conductor is moved in a magnetic field, a voltage will be induced on the conductor. However, current won't flow through t
    11·1 answer
  • Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c
    6·1 answer
  • The heat transfer rate per unit area in a thermal circuit is equivalent to what quantity in an electric circuit? A. voltage B. c
    6·1 answer
  • If a particle moving in a circular path of radius 5 m has a velocity function v = 4t2 m/s, what is the magnitude of its total ac
    15·2 answers
  • Nơi nào có điện tích thì xung quanh điện tích đó có :
    9·1 answer
  • What is it that makes a battery rechargeable? How is it different from a regular battery?
    14·2 answers
  • Determine if the fluid is satisfied​
    10·1 answer
  • An incremental encoder is rotating at 15 rpm. On the wheel there are 40 holes. How many degrees of rotation would 1 pulse be?
    11·1 answer
  • What is applied technology?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!