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4 years ago

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The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction bet

ween this roadway and typical tires is 0.40, what would be the maximum safe driving speed for this horizontal curved section of highway?

1 answer:

kirill [66]4 years ago

8 0

The answer would be 54 m/s as the maximum speed

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For a projectile launched from the vertically from the surface the earth with initial velocity v0, the velocity of the projectil

UNO [17]

Answer:

(dr/dt) = √(v₀² + k²/r - k²/R)

Explanation:

v² = v₀² + k²[(1/r) - (1/R)]

v = velocity of Body launched from the surface of the earth

v = initial velocity of body

k = a constant

r = distance from the centre of the earth

R = radius of the earth

v² = v₀² + k²/r - k²/R

v = √(v₀² + k²/r - k²/R)

But v = dr/dt

(dr/dt) = √(v₀² + k²/r - k²/R)

(dr/√(v₀² + k²/r - k²/R)) = dt

To go one step beyond and integrate the differential equation

∫ (dr/√(v₀² + k²/r - k²/R)) = ∫ dt

Integrating the left hand side from 0 to r and the right hand side from 0 to t

Note, v₀, k and R are all constants

(- 4r²/k²)[√(v₀² + k²/r - k²/R)] = t

√(v₀² + k²/r - k²/R) = (- k² t/4r²)

(v₀² + k²/r - k²/R) = (- k² t/4r²)²

(v₀² + k²/r - k²/R) = (k⁴t²/16r⁴)

(k⁴t²/16r⁴) + (k²/r) = [v₀² - (k²/R)]

k² [(k²t²/16r⁴) + (1/r)] = [v₀² - (k²/R)]

[(k²t²/16r⁴) + (1/r)] = [(v₀²/k²) - (1/R)]

3 0

3 years ago

A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.

belka [17]

Answer:

Approximately $13\; {\rm N \cdot m^{-1}}$ (assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .)

Explanation:

Let $F_{\text{s}}$ denote the force that this spring exerts on the object. Let $x$ denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant $k$ of this spring would ensure that $F_\text{s} = -k\, x$ .

Note that the mass of the object attached to this spring is $m = 540\; {\rm g} = 0.540\; {\rm kg}$ . Thus, the weight of this object would be $m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}$ .

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, $F_{\text{s}} = 5.230\; {\rm N}$ .

The spring in this question was stretched downward from its equilibrium by:

$\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}$ .

(Note that $x$ is negative since this displacement points downwards.)

Rearrange Hooke's Law to find $k$ in terms of $F_{\text{s}}$ and $x$ :

$\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}$ .

3 0

2 years ago

The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the d

tamaranim1 [39]

Answer:

$26852726.19\ \text{N}$

$57.52^{\circ}$

Explanation:

r = Radius of circle = 7 m

w = Width of dam = 60 m

h = Height of the dam will be half the radius = $\dfrac{r}{2}$

A = Area = $rw$

V = Volume = $w\dfrac{\pi r^2}{4}$

Horizontal force is given by

$F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}$

Vertical force is given by

$F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}$

Resultant force is

$F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}$

The hydrostatic force on the dam is $26852726.19\ \text{N}$ .

The direction is given by

$\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}$

The line of action is $57.52^{\circ}$ .

7 0

3 years ago

A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22.0-min rest stop. If the person’s average spee

DanielleElmas [232]

Answer:

(a): It spends on the trip 2.75 hr (2 hours 45 minutes)

(b): The person travels 213.9 km.

Explanation:

Vp= 77.8 km/h

V1= 89.5 km/h

V2= 0 km/h

t2= 0.36hr

t1= ?

Vp= (V1*t1 + V2*t2) / (t1 + t2)

clearing t1:

t1= 2.39 hr

Trip time= t1+t2

Trip time= 2.75 hr (a)

Distance traveled= V1*t1

Distace traveled= 213.9 km (b)

7 0

3 years ago

Is walking just a constant state of falling?

atroni [7]

Answer:

Yes. Walking is controlled falling because you need to let go in order to move forward. If you never let your foot fall, your movements would be stilted and robotic. And according to medical engineers," When we walk normally we are constantly correcting tiny falls to keep ourselves stable."

Explanation:

8 0

4 years ago