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2 years ago

Which of the following statements are true? For the false statements, explain why you think they are false:

Engineering

1 answer:

jarptica [38.1K]2 years ago

8 0

Answer:

Options: (a) and (d)

Explanation:

Measured variable is required for feedback and feed forward as the corrective measures taken by the controller are based on the measured variable as the without the measured variable there won't be any error.

Since, error is the difference between the measured variable and a set standard.

In the feedback control mechanism, evaluation of error is done and then corrective measures are taken according to the error evaluated no matter known or unknown.

Thus option (a) and (d) are correct.

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A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting

sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

F dx = m v dv

We replace and integrate

-β ∫ x³ dx = m ∫ v dv

β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

x_max⁴ = 2 m /β v₀²

Let's look for the speed that the train can have for maximum compression

x_max = 20 cm = 0.20 m

v₀ =√(β/2m) x_max²

Let's calculate

v₀ = √(640 106/2 7.8 104) 0.20²

v₀ = 64.05 0.04

v₀ = 2,562 m / s

v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

v₀ = 9.2 km / h

5 0

2 years ago

An automobile having a mass of 900 kg initially moves along a level highway at 100 km/h relative to the highway. It then climbs

creativ13 [48]

Answer:

ΔKE=-347.278 kJ

ΔPE= 441.45 kJ

Explanation:

given:

mass m=900 kg

the gravitational acceleration g=9.81 m/s^2

the initial velocity $V_{1}$ =100 km/h-->100*10^3/3600=27.78 m/s

height above the highway h=50 m

h1=0m

the final velocity $V_{F}$ =0 m/s

To find:

the change in kinetic energy ΔKE

the change in potential energy ΔPE

assumption:

We take the highway as a datum

solution:

ΔKE=5*m*( $V_{F}$ ^2- $V_{1}$ ^2)

=-347.278 kJ

ΔPE=m*g*(h-h1)

= 441.45 kJ

5 0

2 years ago

What are the three most common metals used in die casting?

Anit [1.1K]

Answer:

Aluminium,Copper,Magnesium

Explanation:

The three most common metal of die casting are as follows

1.Aluminium

2.Copper

3.Magnesium

Die casting is the process in which metal is forced in the die to produces the desired casting product.Generally two type of machines are used like cold chamber and hot chamber machining,it depends on the metals.Die casting produces simple shape of casting ,it can not use for complex casting.

3 0

2 years ago

During an experiment conducted in a room at 25°C, a laboratory assistant measures that a refrigerator that draws 2 kW of power h

zvonat [6]

Answer:

Not reasonable.

Explanation:

To solve this problem it is necessary to take into account the concepts related to the performance of a reversible refrigerator. The coefficient of performance is basically defined as the ratio between the heating or cooling provided and the electricity consumed. The higher coefficients are equivalent to lower operating costs. The coefficient can be greater than 1, because it is a percentage of the output: losses, other than the thermal efficiency ratio: input energy. For a reversible refrigerator the coefficient is given by

$COP_{R,rev} = \frac{1}{\frac{T_1}{T_2}-1}$

Where,

$T_1 =$ High temperature

$T_2 =$ Low Temperature

With our values previous given we can find it:

$T_2 = -30\°C = (-30+273)$

$T_2 = 243K$

$T_1 = 25\°C = (25+273)$

$T_1 = 298K$

With these values we can now calculate the coefficient of performance:

$COP_{R,rev} = \frac{1}{\frac{298}{243}-1}$

$COP_{R,rev} = 4.42$

At the same time we can calculate the work consumption of the refrigerator, this is

$W = \dot{W}\Delta t$

Where,

$\dot{W} =$ Required power input

t = time to remove heat from a cool to water medium

$W = 2kJ/s * 20 min$

$W = 2kJ/s * 1200s$

$W = 2400kJ$

In this way we can calculate the coefficient of the refrigerator directly:

$COP_R = \frac{Q_L}{W}$

Where,

Q = Amoun of heat rejected

$COP_R = \frac{30000}{2400}$

$COP_R = 12.5$

Comparing the values of both coefficients we have that the experiments are NOT reasonable, because the coefficient of a refrigerator is high compared to coefficient of reversible refrigerator.

5 0

2 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

ddd [48]

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

note:

find the attached graph

6 0

2 years ago