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AleksandrR [38]

3 years ago

7

You are driving on a road where the speed limit is 35 mph. If you want to make a turn, you must start to signal at least _______

_ before you turn.
The options are as follows:
A. 10 feet
B. 100 feet
C. 75 feet
D. 20 feet

1 answer:

stich3 [128]3 years ago

3 0

I believe it’s D) 20 feet

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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

simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$

We get the inertia as follows

$I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}$

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

$y=\frac{h}{2} =\frac{0.3m}{2}=0.15m$

If M = Mmax, we have

$(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}$

8 0

3 years ago

Give two causes that can result in surface cracking on extruded products.

Andreas93 [3]

Answer:

1. High friction

2. High extrusion temperature

Explanation:

Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.

Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.

Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.

6 0

3 years ago

Suppose the loop is moving toward the solenoid (to the right). Will current flow through the loop down the front, up the front,

Tems11 [23]

Answer:

See explanation

Explanation:

The magnetic force is

F = qvB sin θ

We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields

F

=

(

20

×

10

−

9

C

)

(

10

m/s

)

(

5

×

10

−

5

T

)

=

1

×

10

−

11

(

C

⋅

m/s

)

(

N

C

⋅

m/s

)

=

1

×

10

−

11

N

6 0

3 years ago

Read 2 more answers

Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t

Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

end

%This brings an end to the program

4 0

3 years ago