Zinc can not be changed into a simpler substance by a chemical or physical process. Therefore, zinc is classified as:

What is your design for an efficient commercial vehicle chassis that incorporates as many commercial truck components into its s

Crank

The design that would be used to make an efficient commercial vehicle chassis that has many realistic truck components is:

Strength
Adequate bending stiffness
Maximum stress
Deflection, etc

<h3>What is a Chassis Frame?</h3>

This refers to the main vehicle structure of a car that bears all the main stress that other components are attached to.

Hence, we can see that to make a good design for an efficient commercial vehicle chassis that has many realistic truck components, we would have to consider various factors such as handling maximum stress, maximum equilateral stress, and deflection.

Read more about chassis frame here:

brainly.com/question/15382293

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5 0

2 years ago

At which location could you place the north pole of a bar magnet so that it would be pushed away from the magnet shown?

Gala2k [10]

It’s a because b makes more since shown as the magnet

7 0

3 years ago

Read 2 more answers

A box of mass 3.6 kg is lifted 5.4 m above the

horsena [70]

So, the energy change that occurs is 190.512 J.

<h3>Introduction</h3>

Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy ( $\sf{\Delta PE}$ ) can be realized in the equation :

$\sf{W = \Delta PE}$

$\sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2}$

$\boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}}$

With the following condition :

W = work of subject (J)
$\sf{\Delta PE}$ = change of potential energy (J)
m = mass (kg)
g = acceleration of the gravity (m/s²)
$\sf{h_2}$ = final height (m)
$\sf{h_1}$ = initial height (m)

<h3>Problem Solving</h3>

We know that :

m = mass = 3.6 kg
g = acceleration of the gravity = 9.8 m/s²
$\sf{h_2}$ = final height = 5.4 m
$\sf{h_1}$ = initial height = 0 m

What was asked :

W = work of subject = ... J

Step by step :

$\sf{W = \Delta PE}$

$\sf{W = m \cdot g \cdot (h_2 - h_1)}$

$\sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)}$

$\sf{W = 3.6 \cdot 9.8 \cdot 5,4}$

$\boxed{\sf{W = 190.512 \: J}}$

So, the energy change that occurs is 190.512 J.

7 0

3 years ago

A proton with an initial speed of 400,000 m/s is brought to rest by an electric field. Did the proton move into a region of high

Anton [14]

A proton is initially moving at high speed

$v = 400000 m/s$

so here initially proton is having high speed and high kinetic energy

now this proton comes to rest by external electric field

so the potential energy of proton will increase or we can say that kinetic energy will convert into electrostatic potential energy

so here the proton must have to move at higher potential region

so here correct answer must be

b. Because the proton is a positive charge and it slows down as it travels, it must be moving from a region of lower potential to a region of higher potential.

8 0

3 years ago

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

So

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

4 0

4 years ago