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4 years ago

What properties of titanium make it attractive for use in race-car and jet-engine components?

1 answer:

3 0

Titanium's high quality to weight proportion and corrosion protection at room and hoisted temperature makes it appealing for use in elite applications. High cost of titanium is the key purpose behind not utilizing it in traveler autos. The cost of large scale manufacturing of the parts would drive the last items cost fundamentally.

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One mole of a gas is placed in a closed system with a 20 L vessel initially at T = 300 K. The vessel is then isothermally expand

Elden [556K]

Answer:

Given that

P = RT/V + a/V²

We know that

H= U + PV

For T= Constant (ΔU=0)

ΔH= ΔU +Δ( PV)

ΔH= Δ( PV)

P = RT/V + a/V²

P V= RT + a/V

dH/dV = d(RT + a/V)/dV

dH/dV = - a/V²

So the expression of dH/dV

$\dfrac{dH}{dV}=\dfrac{-a}{V^2}$

In isothermal process

$\Delta H=nRT\ln{\dfrac{V_2}{V_1}}$ (ΔU=0)

Now by putting the all values

$\Delta H=nRT\ln{\dfrac{V_2}{V_1}}$

$\Delta H=1\times 0.08206\times 300\ln{\dfrac{40}{20}}$

ΔH = 17.06 L.atm

8 0

3 years ago

only 90% can dothis plz helpme i will give brainliest brainliest brainliest brainliest brainliest brainliest brainliest brainlie

stepan [7]

The force exerted by a magnet is called The push or pull of magnetism can act at a distance, which means that the magnet does not have to touch an object to exert a force on it. ... In fact, magnetism is the result of a moving electric charge.Explanation:

8 0

3 years ago

Read 2 more answers

The weight of an object is the same on two different planets. The mass of planet A is only sixty percent that of planet B. Find

natka813 [3]

Answer:

0.775

Explanation:

The weight of an object on a planet is equal to the gravitational force exerted by the planet on the object:

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

For planet A, the weight of the object is

$F_A=G\frac{M_Am}{R_A^2}$

For planet B,

$F_B=G\frac{M_Bm}{R_B^2}$

We also know that the weight of the object on the two planets is the same, so

$F_A = F_B$

So we can write

$G\frac{M_Am}{R_A^2} = G\frac{M_Bm}{R_B^2}$

We also know that the mass of planet A is only sixty percent that of planet B, so

$M_A = 0.60 M_B$

Substituting,

$G\frac{0.60 M_Bm}{R_A^2} = G\frac{M_Bm}{R_B^2}$

Now we can elimanate G, MB and m from the equation, and we get

$\frac{0.60}{R_A^2}=\frac{1}{R_B^2}$

So the ratio between the radii of the two planets is

$\frac{R_A}{R_B}=\sqrt{0.60}=0.775$

6 0

3 years ago

Sam is pulling a box up to the second story of his apartment via a string. The box weighs 92.6 kg and starts from rest on the gr

grin007 [14]

Answer: 2.7 seconds

Explanation:

We only want to answer: How long until the box reaches a height of 13.1 m?

Then we only must integrate the movement equations.

We know that the velocity of the box increases by 3.6 m/s every second, then the acceleration is constant, and can be written as:

a(t) = 3.6m/s^2

Now, for the velocity, we should integrate over time, and because we know that the box starts from rest, the initial velocity (the constant of integration) will be zero.

v(t) = (3.6m/s^2)*t

For the position equation we should integrate again over time, and if we define the position 0 as the ground, we know that the box starts at the ground, then the initial position (the constant of integration) will be zero.

p(t) = (1/2)*(3.6m/s^2)*t^2.

Now we want to find how long will take until the height of the box is equal to 13.1m

Then we must solve:

p(t) = 13.1m = (1/2)*(3.6m/s^2)*t^2

Let's solve this for t.

13.1m = (1/2)*(3.6m/s^2)*t^2

13.1m*2 = (3.6m/s^2)*t^2

26.2m/(3.6m/s^2) = t^2

7.27<u>7</u>.... s^2 = t^2

√(7.27<u>7</u>.... s^2) = t = 2.7 seconds.

So it will take 2.7 seconds for the box to reach the height of 13.1m

4 0

3 years ago

In which direction does the magnetic field in the center of the coil point?

neonofarm [45]

left side magnetic field in the centre of the coil point

6 0

3 years ago