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3 years ago

In Florida a curb indicates and area where parking is prohibited

2 answers:

8 0

Curb striping or painting is used in drop-off and pick-up zones to clarify parking and other curb use rules. The color painted on curbs means: White (or no color): Parking allowed, unless restricted or limited by signs. Blue: Parking for the disabled only.

oksian1 [2.3K]3 years ago

8 0

Answer:

A curb is an Area generally not meant for parking with exceptions

Explanation:

A curb is is like a restraint placed on something as with roads in Florida, In Florida there are different colors used to differentiate the types of curbs found on the road and they are:

Yellow : this is used to indicate an area where parking is strictly forbidden like at road intersections or places that parking might pose a Traffic threat or environmental hazard.
Red : this indicates that no form of stopping, parking or standing is permitted in that area
blue : areas marked with blue are areas meant for disabled persons to park or their driver might park there for them
white : this signifies that the area is not for parking but to drop off passenger or mails
green : signifies parking for a limited time. parking more than the limited time might attract a fine

A curb is area that parking is generally not permitted with exceptions as seen with the different colors used to mark the curbs.

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3 years ago

The uniform beam is supported by two rods AB and CD that have crosssectional areas of 10 mm2 and 15 mm2 , respectively. Determin

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Answer:

w=2.25

Explanation:

It is necessary to determine the maximum w so that the normal stress in the AB and CD rods does not exceed the permitted normal stress.

The surface of the cross-section of the stapes was determined:

A_ab= 10 mm^2

A-cd= 15 mm^2

The maximum load is determined from the condition that the normal stresses is not higher than the permitted normal stress σ_allow.

σ_ab = F_ab/A_ab $\leq$ σ_allow

σ_cd = F_cd/A_cd $\leq$ σ_allow

In the next step we will determine the static size: Picture b).

We apply the conditions of equilibrium:

∑F_x=0

∑F_y=0

∑M=0

∑M_a=0 ==> -w*6*0.5*6*0.75*F_cd*6 =0

==> F_cd = 2*w*k*N

∑F_y=0 ==> F_cd+F_ab - 6*w*0.5 ==>2*w+F_ab -6*w*0.5 =0

==> F_ab = w*k*N

Now we determine the load w

Sector AB:

σ_ab = F_ab/A_ab $\leq$ σ_allow=300 KPa

= w/10*10^-6 $\leq$ σ_allow=300 KPa

w_ab = 3*10^-3 kN/m

Sector CD:

σ_cd = F_cd/A_cd $\leq$ σ_allow=300 KPa

= 2*w/15*10^-6 $\leq$ σ_allow=300 KPa

w_cd = 2.25*10^-3 kN/m

w=min{w_ab;w_cd} ==> w=min{3*10^-3;2.25*10^-3}

==> w=2.25 * 10^-3 kN/m

The solution is:

w=2.25 N/m

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3 years ago

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Answer:

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