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3 years ago

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

Engineering

1 answer:

marta [7]3 years ago

8 0

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

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Oil (SAE 30) at 15.6 oC flows steadily between fixed, horizontal, parallel plates. The pressure drop per unit length along the c

Nitella [24]

Answer:

(a) The volume rate of flow per meter width = 5.6*10⁻³ m²/s

(b) The shear stress acting on the bottom plate = 157.5 N/m²

Explanation:

(a)

Calculating the distance of plate from centre line using the formula;

h = d/2

where h = distance of plate

d = diameter of flow = 9 mm

Substituting, we have;

h = 9/2

= 4.5 mm = 4.5*10^-3 m

Calculating the volume flow rate using the formula;

Q = (2h³/3μ)* (Δp/L)

Where;

Q = volume flow rate

h = distance of plate = 4.5*10^-3 m

μ = dynamic viscosity = 0.38 N.s/m²

(Δp/L) = Pressure drop per unit length = 35 kPa/m = 35000 Pa

Substituting into the equation, we have;

Q = (2*0.0045³/3*0.38) *(35000)

= (1.8225*10⁻⁷/1.14) * (35000)

= 1.60*10⁻⁷ * 35000

= 5.6*10⁻³ m²/s

Therefore, the volume flow rate = 5.6*10⁻³ m³/s

(b) Calculating the shear stress acting at the bottom plate using the formula;

τ = h*(Δp/L)

= 0.0045* 35000

= 157.5 N/m²

u(max) = h²/2μ)* (Δp/L)

= (0.0045²/2*0.38) * 35000

=2.664*10⁻⁵ *35000

= 0.93 m/s

7 0

3 years ago

The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse

sertanlavr [38]

Answer:

T = 5416.67 N

T = -2083.5 N

T = 0

Explanation:

Forward thrust has positive values and reverse thrust has negative values.

part a

Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s

The thrust force represents the horizontal or x-component of momentum equation:

$T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N$

Answer: The thrust force T = 5416.67 N

part b

Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:

$T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N$

Answer: The thrust force T = -2083.5 N reverse direction

part c

Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.

Answer: T = 0 N

5 0

3 years ago

A furnace wall is to be built of 20-cm firebrick and building (structural) brick of same thickness. The thermal conductivities o

Norma-Jean [14]

Answer:

q=2313.04 $W/m^2$

T=690.86°C

Explanation:

Given that

Thickness t= 20 cm

Thermal conductivity of firebrick= 1.6 W/m.K

Thermal conductivity of structural brick= 0.7 W/m.K

Inner temperature of firebrick=980°C

Outer temperature of structural brick =30°C

We know that thermal resistance

$R=\dfrac{t}{KA}$

These are connect in series

$R=\left(\dfrac{t}{KA}\right)_{fire}+\left(\dfrac{t}{KA}\right)_{struc}$

$R=\dfrac{0.2}{1.6A}+\dfrac{0.2}{0.7A}\ K/W$

$R=\dfrac{23}{56A}\ K/W$

Heat transfer

$Q=\dfrac{\Delta T}{R}$

$Q=56A\times \dfrac{980-30}{23}\ W$

So heat flux

q=2313.04 $W/m^2$

Lets temperature between interface is T

Now by equating heat in both bricks

$\dfrac{980-T}{\dfrac{0.2}{1.6A}}=\dfrac{T-30}{\dfrac{0.2}{0.7A}}$

So T=690.86°C

6 0

3 years ago

A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in

blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

Part (A) The stress on the steel wire;

δ = F/A

= 270 / 0.0144

δ = 18750 lb/in² = 19,000 Psi

Part (B) The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

Part (C) The modulus of elasticity of the steel

E = δ/σ

= 19,000 / 0.00063

E = 30,000,000 Psi

4 0

3 years ago

Who here is a genius?

Pachacha [2.7K]

Answer:

im kinda smart whyy?

Explanation:

3 0

3 years ago

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