Which statement is supported by the information in the diagram and table below?

4. The bar has cross-sectional area A and modulus of elasticity E. If an axial force F directed toward the right is applied at C

aniked [119]

Answer:

a) ΔL/L = F / (E A), b) $L_{f}$ = L (1 + L F /(EA) )

Explanation:

Let's write the formula for Young's module

E = P / (ΔL / L)

Let's rewrite the formula, to have the pressure alone

P = E ΔL / L

The pressure is defined as

P = F / A

Let's replace

F / A = E ΔL / L

F = E A ΔL / L

ΔL / L = F / (E A)

b) To calculate the elongation we must have the variation of the length, so the length of the bar must be a fact. Let's clear

ΔL = L [F / EA]

$L_{f}$ -L = L (F / EA)

$L_{f}$ = L + L (F / EA)

$L_{f}$ = L (1 + L (F / EA))

4 0

2 years ago

Pressure can affect the density of Solids Liquids Gases Metals

gtnhenbr [62]

Answer;

-Gases

Explanation;

-Pressure can affect the density of gases.

-Density of gases changes with pressure and temperature because gases are compressible fluid and because they are compressible, when pressure increases molecules come closer to each other which means increase in density and when pressure drops molecules of gases become free to expand and get away from each other which density decrease.

7 0

2 years ago

how many kilograms off mercury would fill a 5litre container if the density of mercury is 13.6grams per cm3

Sidana [21]

Answer:

68kg

Explanation:

1 cm^3 is the same as 1 mL and there are 5000mL in 5L

Therefore if the density is 13.6g/mL we multiply 13.6 by 5000 to get the amount of grams required = 68000g which is 68kg

7 0

2 years ago

By what percent is the torque of a motor decreased if its permanent magnets lose 6.5 % of their strength

djyliett [7]

6 is the answer I remember the answer from when I took this and it was easy

4 0

1 year ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

2 years ago