Just some random stufff

A 993-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen

tino4ka555 [31]

Answer:

13177.34 J

Explanation:

Work done = force × distance

work done by the engine = kinetic energy + potential energy + work done friction

kinetic energy due to the car's speed = 1/2mv² = 4468.5 J

potential energy due to the height = mgh = 993 kg × 9.8 m/s² × 0.6 m = 5838.84 J

work done by friction = 2870 J

work done by engine = 5838.84 J + 2870 J + 4468.5 J = 13177.34 J

7 0

3 years ago

What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5 × 10-4

White raven [17]

Answer:

The question is a problem that requires the principles of fracture mechanics.

and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.

Explanation:

Max Stress, σ = 2σ₀√(α/ρ)

where,

σ₀ = Tensile stress = 190MPa = 1.9x10⁸Pa

α = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m

ρ = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m

Max Stress, σ = 2 x 1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵

Max Stress, σ = 2 x 1.9x10⁸ x 6.708 Pa

Max Stress, σ = 2549MPa

Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa

7 0

4 years ago

Consider a 0.15-mm-diameter air bubble in a liquid. Determine the pressure difference between the inside and outside of the air

777dan777 [17]

Answer:

a) $\Delta P= 2666.66 Pa$

b) $\Delta P= 3200 Pa$

Explanation:

Given that

Diameter ,d= 0.15 mm

We know that pressure difference is given as

$\Delta P=\dfrac{4\sigma }{d}$

Now by putting the values

When surface tension 0.1 N/m :

The surface tension , $\sigma=0.1\ N/m$

$\Delta P=\dfrac{4\times 0.1 }{0.15}\times 10^3\ Pa$

$\Delta P= 2666.66 Pa$

When surface tension 0.12 N/m :

The surface tension , $\sigma=0.12\ N/m$

$\Delta P=\dfrac{4\times 0.12 }{0.15}\times 10^3\ Pa$

$\Delta P= 3200 Pa$

a) $\Delta P= 2666.66 Pa$

b) $\Delta P= 3200 Pa$

4 0

4 years ago

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 3.5 mm (0.14 in

RideAnS [48]

To resolve this problem we have,

$R=3.5mm\\F_f1=950N\\L_1=50mm\\b=12mm\\L_2=40mm$

$F_{f2}$ is unknown.

With these dates we can calculate the Flexural strenght of the specimen,

$\sigma{fs}=\frac{F_{f1}L}{\pi R^3}\\\sigma{fs}=\frac{(950)(50*10^{-3})}{\pi 3.5*10^{-3}}\\\sigma{fs}=352.65Mpa$

After that, we can calculate the flexural strenght for the square cross section using the previously value.

$\sigma{fs}=\frac{F_{f2}L}{\pi R^3}\\(352.65*10^6)=\frac{3Ff(40*10^{-3})}{2(12*10^{-10})}\\F_{f2}=\frac{352.65*10^6}{34722.22}\\F_{f2}=10156.32N\\F_{f2}=10.2kN$

6 0

3 years ago

A distillation column at 101 kPa is used to separate 350 kmol/h of a bubble point mixture of toluene and benzene into an overhea

AURORKA [14]

Answer:

A)

D = 158.42 kmol/h

B = 191.578 kmol/h

B) Rmin = 1.3095

Explanation:

a) Determine the distillate and bottoms flow rates ( D and B )

F = D + B ----- ( 1 )

Given data :

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

B) Determine the minimum reflux ratio Rmin

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )

q - line ; y = ( 9 / 9-1 ) x - xf/9-1

therefore ; x = 0.45

Finally Rmin

= (( 0.97 / (Rm + 1 )) = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

8 0

3 years ago