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3 years ago

What are the causes and solutions of social problems ?what are the causes and solutions of social problems

Engineering

1 answer:

aksik [14]3 years ago

7 0

Answer:

causes: unemployment, poverty,Lack of education,urbanization e.t.c

solutions:making people educated, giving people jobs

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Which of the following machine parts always require

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Answer:fan blades, rotating , abrasive wheel machinery

Explanation:

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3 years ago

Concrete ___ support and anchor the bottom of steel columns and wood post, which support beams that are pare of framing system o

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I wanna say it’s D post support

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3 years ago

The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular veloc

madreJ [45]

Answer:

Option B

$116 ft/s^{2}$

Explanation:

$\theta=2 rev=2(2\pi)=4\pi$

$\alpha \theta=0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (4\pi)= 0.5(\omega_f^{2}-\omega_i^{2})$

$\alpha (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-\omega_i^{2})$

$(2) (8\pi)= (\omega_f^{2}-4^{2})$

$\omega_f=8.14 rads/s$

$v=r\omega=1.75*8.14=14.245 ft/s$

Centripetal acceleration $=\omega_f^{2} r=8.14^{2}*1.75=115.95 ft/s^{2}$

Tangential component=dr=2*1.75=3.5

$Resultant=\sqrt{3.5^{2}+115.95^{2}}\approx 116 ft/s^{2}$

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3 years ago

All circuits need three basic parts: an energy source, wires, and the object that is going to change the electrical energy into

Radda [10]

load every electric circuit,regardless of where it is or how large or small, has four basic parts: an energy source (ac or dc),a conductor (wire), an electrical load (device), and at least one controller(switch)

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3 years ago

Read 2 more answers

A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft

tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ; f_{Lw}=6.6 mi/h

5' on right ; f_{Lc} = 0.4 mi/hr

3' on left ; no adjustment

3 lanes in each direction f n = 3 mi/h

$v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}$

$f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}$

= 0.877

$v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}$

= 1,555 veh/hr/lane

$FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}$

= (55 + 5) - 6.6 - 0.4 -3 -0

= 50 mi/h

$D = \dfrac{V_P}{s}$

$D = \dfrac{1555}{55} =28.27$

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0

4 years ago

What are the causes and solutions of social problems ?what are the causes and solutions of social problems ​

What are the causes and solutions of social problems ?what are the causes and solutions of social problems