1. You do not need to remove the lead weights inside tires before recycling them.

A specific system has four sensors capable of producing either a 0 or a 1 output. The system functions properly. An alarm must be triggered when two or more sensors output 1. Design the simplest circuit to raise the alarm only when precisely one sensor outputs 1. Repeat this for a system with seven sensors. Hint: Before getting overwhelmed with a 128-row truth table, consider whether a Sum of Products or Product of Sums approach would be more effective.

Answer:

L_f = 26.025 ft

v_f = 51.77 ft/min

Explanation:

Given:

- Initial slab thickness, t_o = 2 in

- Initial slab width, w_o = 10 in

- Initial slab length, L_o = 12.0 ft

- Thickness reduction each step, r = 75%

- Width increase each step, m = 3%

- Initial entry speed vi = 40 ft/min

- Roll speed remains constant

Required:

a) Length

b) Exit velocity

For the final slab

Solution:

- After three passes, the final thickness (t_f) is determined as:

                      t_f =  ( r / 100 )^n * t_o

Where n signifies the number of passes.

                     t_f = ( 75 / 100 ) ^3 * ( 2.0 )

                     t_f = 0.844 in

- The final width after three passes is calculated as:

                      w_f =  ( m / 100 + 1 )^n * w_o

Where n denotes the number of passes.

                     w_f = ( 3 / 100 + 1 ) ^3 * ( 10.0 )

                     w_f = 10.927 in

- Assuming there is no loss in material, the final length of the slab can be obtained:

                    t_o*w_o*L_o = t_f*w_f*L_f

                    L_f = ( 2 * 10 * 12 ) / ( 0.844 * 10.927 )

                    L_f = 26.025 ft

- The equation for volume rate can be applied since the roll speed stays the same. Thus, we equate the conditions before and after the third step:

                   t_i*w_i*v_i = t_f*w_f*v_f

Where v_i is the initial speed at entry:

            v_f signifies the exit speed after the third step.

                   (0.75)^2 * 2 * (1.03)^2 * 10 * 40 =  (0.844)*(10.927)*v_f

                  v_f = 51.77 ft/min    

pressure tactic Response:

Reasoning:

because the child will shame the parents, leading them to give in to the child's demands.

Answer:

0.285 lbm per cubic foot

Explanation:

Length of tank = 50 inches = 50/12 feet = 4.17 feet

Diameter of tank = 16 inches = 16/12 feet = 1.33 feet

Weight of water = 1.65 lbm

Density of water =?

We know the density of a substance is defined as

ρ = w/v

where ρ is the density in lbm per cubic foot

w is the weight in lbm

v is the volume in cubic feet

Volume of a cylinder =

where d is diameter

l is the length

volume = = 5.79 cubic feet

Hence, the density of water will be

ρ = w/v = 1.65/5.79 = 0.285 lbm per cubic foot

Answer: 0.021818m = 2.18x10^-²m

Explanation: According to Laplace's principle

Design pressure equals 12 bar or 1200000N/m².

T denotes the wall thickness.

Design stress is 110Mn/m² equating to 110,000,000N/m².

Radius is calculated as diameter/2 = 4/2 = 2m.

Design stress equates to hoop stress: Pr/t, where p indicates internal pressure, r signifies radius, and t represents wall thickness.

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110,000,000 = 1,200,000 x 2/t.

Solving for t gives t = 2,400,000/110,000,000.

Thus, t = 0.021818m.

Ultimately, t = 2.18x10^-2m.

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