Need help please????????!!!!!!

A civil engineer is asked to design a curved section of roadway that meets the following conditions: With ice on the road, when

lianna [129]

Answer:

1. $3.4^{o}$

2. 163.3 m

Explanation:

Static friction between road and rubber, μs =0.06

The maximum speed of the car, v = 50 km/h

= (50)(1000/3600) m/s

= 13.89 m/s

The acceleration due to gravity, $g = 9.81 m/s^{2}$

The frictional force, f = μsN ...... (1)

The component mg cosθ which balance the normal reaction N

The component mg sinθ acts in an opposite direction to the frictional force f.

ΣF = mg sinθ-f = 0 ...... (2)

Substitute the equation (1) in equation (2), we get

ΣF = mgsinθ-μsN = 0

mgsinθ-μsmgcosθ =0

μs = sinθ/cosθ

tanθ = μs

θ = tan-1( μs) = tan-1(0.06) = $3.4^{o}$

(b)The vertical component of the force is

N cosθ = fsinθ+mg

N cosθ = μsNsinθ+mg

N[cosθ- μs sinθ] = mg ...... (3)

The horizontal component of the force along the motion of the car is

Nsinθ+fcosθ = ma (Centripetal acceleration, $a = \frac {v^{2}}{r}$

Nsinθ+fcosθ = $m(\frac {v^{2}}{r})$

Nsinθ+μsNcosθ = $m(\frac {v^{2}}{r})$

N[sinθ+μs cosθ] = $m(\frac {v^{2}}{r})$ ...... (4)

Dividing the equation (4) with equation (3),

[sinθ+μscosθ]/[cosθ- μs sinθ] = $\frac {v^{2}}{rg}$

cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] = $\frac {v^{2}}{rg}$

[tanθ+μs]/[1-μs tanθ] = $\frac {v^{2}}{rg}$

From part (1), tanθ = μs

Then the above equation becomes

$\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

$\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}$

Therefore, the minimum radius of the curvature of the curve is

$r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}$

= $\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}$

= $\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}$

= 163.3 m

5 0

3 years ago

Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature

Ronch [10]

Answer:

The surface temperature increases when two bodies are rubbed against each other due to friction.

Explanation:

No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.

Friction is a resistance experienced by the two bodies when they are moved against each other.

The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.

Thus, resulting in rise in the surface temperature of the two bodies.

Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.

7 0

3 years ago

A motor driven water pump operates with an inlet pressure of 96 kPa (absolute) and mass flow rate of 120 kg/min. The motor consu

NeX [460]

Answer:

The maximum water pressure at the discharge of the pump (exit) = 496 kPa

Explanation:

The equation expressing the relationship of the power input of a pump can be computed as:

$E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}$

where;

m = mass flow rate = 120 kg/min

the pressure at the inlet $P_1$ = 96 kPa

the pressure at the exit $P_2$ = ???

the pressure $\rho$ = 1000 kg/m³

∴

$0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}$

$0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}$

$\dfrac{800000}{2} = P_2-96000$

400000 = P₂ - 96000

400000 + 96000 = P₂

P₂ = 496000 Pa

P₂ = 496 kPa

Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa

8 0

3 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

3 years ago

W10L1-Show It: Pythagorean Theorem<br> Calculate the total material in the picture.<br> 4<br> 3

Fantom [35]

Answer:

35

Explanation: I really dont even know, I just used up all my tries on it and got it wrong on every other thing i chose. So it's 35 i believe cause its the only answer i didnt choose.

7 0

3 years ago