You sometimes service vehicles with rust damage. It's safe to lift them on a two-post frame-engaging lift without checking the vehicle lifting points. This is false.
<h3>What is rust?</h3>
Rust is an iron oxide, a typically reddish-brown oxide created when iron and oxygen react catalytically with moisture from the air or water. Iron(III) oxide-hydroxide and hydrous iron(III) oxides make up rust, which is often linked to the corrosion of refined iron.
Rusting is the chemical process that causes a red or orange coating to form on the surface of metals. Corrosion incorporates rusting. Steel and iron rust when they come into contact with oxygen and water, which are both necessary for rusting to take place.
It should be noted that when servicing vehicles with rust damage, It's not safe to lift them on a two-post frame-engaging lift without checking the vehicle lifting points. This can cause an accident.
Learn more about rust on:
brainly.com/question/1477535
#SPJ1
You sometimes service vehicles with rust damage. It's safe to lift them on a two-post frame-engaging lift without checking the vehicle lifting points.
True
False
Answer
given,
Speed of vehicle = 65 mi/hr
= 65 x 1.4667 = 95.33 ft/s
e = 0.07 ft/ft
f is the lateral friction, f = 0.11
central angle,Δ = 38°
The PI station is
PI = 250 + 50
= 25050 ft
using super elevation formula



r = 1568 ft
As the road is two lane with width 12 ft
R = 1568 + 12/2
R = 1574 ft
Length of the curve


L = 1044 ft
Tangent of the curve calculation


T = 542 ft
The station PC and PT are
PC = PI - T
PC = 25050 - 542
= 24508 ft
= 245 + 8 ft
PT = PC + L
= 24508 + 1044
=25552
= 255 + 52 ft
the middle ordinate calculation


MO = 85.75 ft
degree of the curvature


D = 3.64°
5.85 psig
Using a specific gravity of 0.75 as an average for red\automobile gasoline.
Water at standard conditions (60 degF) is 2.31 feet = 1 psig
80/2.31 then multiply x .75 to compensate for specific gravity of water being 1.0
Answer:
I = 5.0 A V=0
Explanation:
Assuming that the ammeter is an ideal one (which means that the internal resistance is negligible compared with other resistors in the circuit), as the voltage through the capacitor can't change instantaneously, just after the switch is closed, will behave like a short, so applying Ohm's law, the current in the circuit will be as follows:

As the voltage in the capacitor, can't change instantaneously, assuming an ideal voltmeter (infinite resistance) , the lecture on the voltmeter across the capacitor will be just 0.
As time goes by, the voltage measured will follow the following equation:

We see that when t=0, V=0.