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3 years ago

Why material selection is important to design and manufacturing?

Physics

2 answers:

Darina [25.2K]3 years ago

5 0

Answer:

You want your product to be as strong and as long lasting as possible. There are also the safety implications to consider. You see, dangerous failures arising from poor material selection are still an all too common occurrence in many industries.

Explanation:

Gwar [14]3 years ago

3 0

Answer:

. You want your product to be as strong and as long lasting as possible. There are also the safety implications to consider. You see, dangerous failures arising from poor material selection are still an all too common occurrence in many industries. yep that the answer have a Great day

Explanation:

(◕ᴗ◕✿)

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Which statement about real gases is true? A) Forces of attraction and repulsion exist between gas particles at close range. B) T

olga55 [171]

Answer: A) Forces of attraction and repulsion exist between gas particles at close range.

Explanation:

The <u>Ideal Gas equation</u> is:

$P.V=n.R.T$

Where:

$P$ is the pressure of the gas

$V$ is the volume of the gas

$n$ the number of moles of gas

$R$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them. In this sense, real gases can behave approximately to an ideal gas, under conditions of high temperature and low pressures.

However, at low temperatures or high pressures, real gases deviate significantly from ideal gas behavior. This is because at low temperatures molecules begin to move slower, allowing the repulsive and attractive forces among them to take effect. In fact, <u>the attraction forces are responsible for the condensation of the gas</u>. In addition, at high pressures the volume of molecules cannot be approximated to zero, hence the volume of these molecules is not negligible anymore.

7 0

3 years ago

A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

aleksklad [387]

Answer:

$V=9.2565m/s$

Explanation:

From the question we are told that:

Force $F = 34 N$

Time $t = 0.6 s$

Length of pedal $l_p=16.5cm \approx0.165m$

Radius of wheel $r = 33 cm = 0.33 m$

Moment of inertia, $I = 1200 kgcm2 = 0.12 kg.m2$

Generally the equation for Torque on pedal $\mu$ is mathematically given by

$\mu=F*L\\\mu=34*0.165$

$\mu=5.61N.m$

Generally the equation for angular acceleration $\alpha$ is mathematically given by

$\alpha=\frac{\mu}{l}$

$\alpha=\frac{5.61}{0.12}$

$\alpha=46.75$

Therefore Angular speed is \omega

$\omega=\alpha*t$

$\omega=(46.75)*(0.6)$

$\omega=28.05rad/s$

Generally the equation for Tangential velocity V is mathematically given by

$V=r\omega$

$V=(0.33)(28.05)$

$V=9.2565m/s$

5 0

3 years ago

Must physics always be explained by math?

cluponka [151]

Answer:

Because Physics is mainly composited of math

Explanation:

If you really think about it the only way any science works is off maths, whether that be measuring the amount of NaOH to put into a solution or determening the amount of sugar you want in a cake it all relys on maths

5 0

3 years ago

A 1180kg car is moving at a speed of 85.5 km/h. Find the force needed to bring the car to rest in a distance of 300m

yarga [219]

Answer:

1110 N

Explanation:

First, find the acceleration.

Given:

Δx = 300 m

v₀ = 85.5 km/h = 23.75 m/s

v = 0 m/s

Find: a

v² = v₀² + 2aΔx

(0 m/s)² = (23.75 m/s)² + 2a (300 m)

a = -0.94 m/s²

Find the force:

F = ma

F = (1180 kg) (-0.94 m/s²)

F = -1110 N

The magnitude of the force is 1110 N.

3 0

3 years ago

Cart A of inertia m has attached to its front end a device that explodes when it hits anything, releasing a quantity of energy E

Leviafan [203]

We need to write down momentum and energy conservation laws, this will give us a system of equation that we can solve to get our final answer. On the right-hand side, I will write term after the collision and on the left-hand side, I will write terms before the collision.
Let's start with energy conservation law:
$\frac{mv^2}{2}+\frac{2mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+\frac{2mv_{B}^2}{2}$
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2$
This equation tells us that kinetic energy of two carts before the collision and 3 quarters of explosion energy is beign transfered to kinetic energy of the cart after the collision.
Let's write down momentum conservation law:
$mv+2mv=mv_A+2mv_B\\ 3mv=mv_A+2mv_B\\$
Because both carts have the same mass we can cancel those out:
$3v=v_A+2v_B$
Now we have our system of equation that we have to solve:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\ 3v=v_A+2v_B$
Part A
We need to solve our system for $v_a$ . We will solve second equation for $v_b$ and then plug that in the first equation.
$3v=v_A+2v_B\\ 3v-v_A=2v_B\\ v_B=\frac{3v-v_A}{2}$
Now we have to plug this in the first equation:
$\frac{3mv^2}{2}+0.75E=\frac{mv_{A}^2}{2}+mv_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
We will multiply the first equation with 2 and divide by m:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_B=\frac{3v-v_A}{2}\\$
Now we plug in the second equation into first one:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+2\frac{(3v-v_A)^2}{4}\\ 3v^2+\frac{3E}{2m}=v_{A}^2}+\frac{9v^2-6v\cdot v_A+v_{A}^2}{2} /\cdot 2\\ 6v^2+\frac{3E}{m}=2v_{A}^2+9v^2-6v\cdot v_A+v_{A}^2}\\ 3v_A^2-6v\cdot v_a+3(v^2-\frac{E}{m})=0/\cdot\frac{1}{3}\\ v_A^2-3v\cdot v_A+ (v^2-\frac{E}{m})=0$
We end up with quadratic equation that we have to solve, I won't solve it by hand.
Coefficients are:
$a=1\\
b=-6v\\
c=v^2-\frac{E}{m}$
Solutions are:
$v_A=\frac{3v+\sqrt{5v^2+\frac{4E}{m}}}{2},\:v_A=\frac{3v-\sqrt{5v^2+\frac{4E}{m}}}{2}$
Part B
We do the same thing here, but we must express $v_a$ from momentum equation:
$3v=v_A+2v_B\\
v_A=3v-2v_B$
Now we plug this into our energy conservation equation:
$3v^2+\frac{3E}{2m}=v_{A}^2}+2v_{B}^2\\v_A={3v-v_B}\\
3v^2+\frac{3E}{2m}=(3v-v_B)^2+2v_B^2\\
3v^2+\frac{3E}{2m}=9v^2-6v\cdot v_B+v_B^2+2v_B^2\\
3v^2+\frac{3E}{2m}=3v_B^2-6v\cdot v_B+9v^2\\
3v_B^2-6v\cdot v_B+9v^2-3v^2-\frac{3E}{2m}=0\\
3v_B^2-6v\cdot v_B+(6v^2-\frac{3E}{2m})=0
$
Again we end up with quadratic equation. Coefficients are:
$a=3\\
b=-6v\\
c=6v^2-\frac{3E}{2m}$
Solutions are:
$v_B=\frac{6v+\sqrt{-36v^2+\frac{18E}{m}}}{6},\:v_B=\frac{6v-\sqrt{-36v^2+\frac{18E}{m}}}{6}$

8 0

3 years ago

Why material selection is important to design and manufacturing?​

Why material selection is important to design and manufacturing?