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3 years ago

The loneliest people are to kindest

Engineering

2 answers:

ryzh [129]3 years ago

6 0

Answer:

these r really good lines

Explanation:

And the first line is the best

valkas [14]3 years ago

5 0

Answer:

The most damaged people are the wisest is a fact

Explanation:

You might be interested in

A train which is traveling at 70 mi/hr applies its brakes as it reaches point A and slows down with a constant deceleration. Its

Ugo [173]

Answer:

a) 0 mi/s^2

b) 52 mi/s

Explanation:

Assuming the crossing is 1/2 mile past point A and that point B is near point A (it isn't clear in the problem)

The train was running at 70 mi/h at point A and with constant deceleration reachesn the crossing 1/2 mile away with a speed of 52 mi/h

The equation for position under constant acceleration is:

X(t) = X0 + V0 * t + 1/2 * a * t^2

I set my reference system so that the train passes point A at t=0 and point A is X = 0, so X0 = 0.

Also the equation for speed under constant acceleration is:

V(t) = V0 + a * t

Replacing

52 = 70 + a * t

Rearranging

a * t = 52 - 70

a = -18/t

I can then calculate the time it will take it to reach the crossing

1/2 * a * t^2 + V0 * t - X(t) = 0

Replacing

1/2 (-18/t) * t^ + 70 * t - 1/2 = 0

-9 * t + 70 * t = 1/2

61 * t = 1/2

t = (1/2)/61 = 0.0082 h = 29.5 s

And the acceleration is:

a = -18/0.0082 = -2195 mi/(h^2)

To beath the train the car must reach the crossing in 29.5 - 4.3 = 25.2 s

X(t) = X0 + V0 * t + 1/2 * a * t^2

52 mi/h = 0.0144 mi/s

1/2 = 0 + 0.0144 * 25.2 + 1/2 * a * 25.2^2

1/2 = 0.363 + 317.5 * a

317.5 * a = 0.5 - 0.363

a = 0.137/317.5 = 0.00043 mi/s^2 (its almost zero)

The car should remain at about constant speed.

It will be running at the same speed.

4 0

3 years ago

Convection ovens operate on the principle of inducing forced convection inside the oven chamber with a fan. A small cake is to b

svetlana [45]

complete question:

attached

Answer:

2356.11 W/m^2

6100.11 W/m^2

Explanation:

Assumptions:

1. Steady-state conditions.

2. The cake is placed in a large surrounding.

3. Heat flux delivered to the cake is due to convection and radiation.

Case 1

Since convection feature is disabled the mode of heat transfer associated with this situation is through free convection and radiation.

q''(free) = [q''(free convection+q''(radiation) ]W/m^2

= h_free(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

= 3 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -

(24+273K)^4 ]

= 468 +1881.11

= 2356.11 W/m^2

Case 2

Since convection feature is enabled or activated the mode of heat transfer associated with this situation is through forced convection and radiation.

q''(free) = [q''(forced convection+q''(radiation) ]W/m^2

= h_forced(T_infinty - T_i) + εσ(T_air^4 - T_i^4)

= 27 W/m^2K(180°C - 24°C) + 0.97*5.67*10^-8*[(180+273K)^4 -

(24+273K)^4 ]

= 4212 +1881.11

= 6100.11 W/m^2

1. The total heat flux is is 2.58 times higher when the convection feature is activated. Therefore the cake will bake faster during this condition.

2. The contribution of convection heat flux under natural(free) convection is very low as compared to the contribution during forced convection.

3. The heat transfer due to radiation is same in both the cases.

4. Only 19.9 % of the total heat flux is contributed by free convection in the first case.

5. In the second case 69 % of the total heat flux is contributed by forced convection.

5 0

3 years ago

-Electronic control modules can easily evaluate the voltage and current levels of circuits to which they are connected and deter

erma4kov [3.2K]

Answer:

multiplexing

Explanation:

3 0

3 years ago

A new 1.2-ha suburban residential development is to be drained by a storm sewer that connects to the municipal drainage system.

valentinak56 [21]

Answer:

Time of concentration, $t_d$ ⇒ 1280 min

Peak runoff rate, $Q_p$ ⇒4.185 ff³/s

Explanation:

See detailed explanation

6 0

3 years ago

A 50-kN hydraulic press performs pressing and clamping actions. The clamping cylinder force is 4 kN. The pressing cylinder strok

galben [10]

Answer:

The attached figure shows the hydraulic circuit using one sequence valve to control two simultaneous operations performed in proper sequence in one direction only. In the other direction, both the operations are simultaneous.

When we keep the 4/2 DCV in crossed arrow position, oil under pressure is supplied to the inlet port of the sequence valve. It directly flows to Head end port-1. Hence Cylinder 'C1' extends first.

By the end of the extension of cylinder 'C1', pressure in the line increases and hence poppet of sequence valve is lifted off from its seat and allows oil to flow to port-2 and hence, Cylinder 'C2 extends completing the pressing operation.

In the straight-arrow position of 4/2 DCV the oil under pressure reaches the rod end of both the cylinders C1 and C2 simultaneously through port-3. This causes both the cylinders to retract simultaneously.

Also, a Flow control valve is provided tho control the velocity of clamping

Explanation:

find attached the figure

4 0

3 years ago