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2 years ago

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The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

s of curvature r r, in feet. A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet. What is the maximum safe speed?

1 answer:

mel-nik [20]2 years ago

4 0

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

$v=\sqrt{2.5\ r}$ .........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

$v=\sqrt{2.5\ \times 380}$

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

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Which type of circuit is shown?

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Answer:

I think it's D. open series circuit .

Explanation:

<em>hope</em><em> it</em><em> helps</em><em> you</em><em>!</em>

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2 years ago

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

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2 years ago

The interior space of large box is kept at 30 C. The walls of the box are 3 m high and have a ‘sandwich’ construction consisting

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Answer:

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$

$T_1=27.58\ ^{\circ}C$ & $T_2=2.41875\ ^{\circ}C$

Explanation:

Given:

interior temperature of box, $T_i=30^{\circ}C$

height of the walls of box, $h=3\ m$

thickness of each layer of bi-layered plywood, $x_p=1.25\ cm=0.0125\ m$

thermal conductivity of plywood, $k_p=0.104\ W.m^{-1}.K^{-1}$

thickness of sandwiched Styrofoam, $x_s=5\ cm=0.05\ m$

thermal conductivity of Styrofoam, $k_s=0.04\ W.m^{-1}.K^{-1}$

exterior temperature, $T_o=0^{\circ}C$

<u>From the Fourier's law of conduction:</u>

$\dot Q=\frac{dT}{(\frac{x}{kA}) }$

$\dot Q=\frac{dT}{R_{th} }$ ....................................(1)

<u>Now calculating the equivalent thermal resistance for conductivity using electrical analogy:</u>

$R_{th}=R_p+R_s+R_p$

$R_{th}=\frac{x_p}{k_p.A}+\frac{x_s}{k_s.A}+\frac{x_p}{k_p.A}$

$R_{th}=\frac{1}{A} (\frac{x_p}{k_p}+\frac{x_s}{k_s}+\frac{x_p}{k_p})$

$R_{th}=\frac{1}{A} (\frac{0.0125}{0.104}+\frac{0.05}{0.04}+\frac{0.0125}{0.104})$

$R_{th}=\frac{1.4904}{A}$ .....................(2)

Putting the value from (2) into (1):

$\dot Q=\frac{30-0}{\frac{1.4904}{A} }$

$\dot Q=\frac{30\ A}{1.4904}$

$\frac{\dot Q}{A} =20.129\ W.m^{-2}$ is the heat per unit area of the wall.

The heat flux remains constant because the area is constant.

<u>For plywood-Styrofoam interface from inside:</u>

$\frac{\dot Q}{A} =k_p.\frac{T_i-T_1}{x_p}$

$20.129=0.104\times \frac{30-T_1}{0.0125}$

$T_1=27.58\ ^{\circ}C$

&<u>For Styrofoam-plywood interface from inside:</u>

$\frac{\dot Q}{A} =k_s.\frac{T_1-T_2}{x_s}$

$20.129=0.04\times \frac{27.58-T_2}{0.05}$

$T_2=2.41875\ ^{\circ}C$

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2 years ago

A 0.105- kg hockey puck moving at 24 m/s is caught and held by a 75-kg goalie at rest. With what speed does the goalie slide on

Veseljchak [2.6K]

Answer:

<u><em>0.03 m/s</em></u>

Explanation:

<em>Applying law of conservation of momentum, </em>

<em>m₁v₁ + m₂v₂ = (m₁ + m₂)v</em>
<em>0.105(24) + 75(0) = (0.105 + 75)v</em>
<em>75.105v = 2.52</em>
<em>v = 2.52/75.105</em>
<em>v = </em><u><em>0.03 m/s</em></u>

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1 year ago

The potential difference between points A and B in an electric

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Answer:

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2 years ago