What limitations exist when attempting to categorize hair by race?

A new interstate highway is being built with a design speed of 120 km/h. For one of the horizontal curves, the radius (measured

nikklg [1K]

Answer:

28.79%

Explanation:

Given

Design Speed, V = 120km/h = 33.33m/s

Radius, R = 300m

Side Friction, Fs = 0.09

Gravitational Constant = 9.8m/s²

Using the following formula, we'll solve the required rate of superelevation.

e + Fs = V²/gR where e = rate

e = V²/gR - Fs

e = (33.33)²/(9.8 * 300) - 0.09

e = 0.287853367346938

e = 28.79%

Hence, the required rate of superelevation for the curve is calculated as 28.79%

4 0

3 years ago

Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B

pogonyaev

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

8 0

3 years ago

An inverse-time circuit breaker (CB) is used for branch-circuit short-circuit and ground-fault protection for a 30-horsepower, 2

Maru [420]

Answer:

Explanation:

Motor rating is given in horsepower (hp), it will be converted in watt (W).

Standard to install circuit breaker for an electric circuit is usually 20% ~ 25% more than the Rated Current of the circuit

while

Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

So, to find the maximum capacity of the circuit breaker, rated current of the motor will be multiplied by 1.2 ~ 1.25

Step by Step Explanation:

30hp = 22371W (as 1hp = 745.7)

Assuming unity power factor (cosФ=1) and 208V phase to phase voltage:

Rated Power (watt) = √3 . V.I. cosФ

{if 208V is phase to neutral voltage, then use following formula:

Rated Power (watt) = 3 . V.I. cosФ}

$\frac{22371}{\sqrt{3} * 208 * 1} = I$

Rated Current = 62.169A

So, required maximum rating for circuit breaker is:

20% to 25% of the rated current = 62.17*1.2 ~ 62.17*1.25

=74.6A ~ 77.7A

Hence, any breaker between the above mentioned rating will be appropriate.

3 0

3 years ago

If it takes 200 joules of energy to lift a bucket of water 3 meters in 2 seconds, how much power would be required to do the sam

Harlamova29_29 [7]

200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.

Power = (work or energy) / (time to do the work or move the energy)

Power = (200 joules) / (5 seconds)

Power = 40 watts

3 0

3 years ago

A car of mass m accelerates from rest along a level straight track, not at constant acceleration but with constant engine power,

lisabon 2012 [21]

Because P = W ÷ t, and W = F*t, you can substitute (W) for (F*t). Then substitute (F) for (m*a). This will leave you with P = (m*a*d)/t. Since you need velocity, youd want to solve for a so you can use v = a*t. a = (P*t)/(m*d) therefore, substituting a in v = a*t, v = (P*t*t)/(m*d)

4 0

3 years ago