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2 years ago

13

Tech A states that friction brakes lose energy as heat. Tech B states that a brake-by-wire system converts this energy into usef

ul work. Who is
correct?

1 answer:

hjlf2 years ago

4 0

Tech A.......................

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Your local hospital is considering the following solution options to address the issues of congestion and equipment failures at

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2 years ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

2 years ago

List and explain 4 factors you need to observe while stick welding to make a good “consistent” bead

ch4aika [34]

Answer:I don’t know this one

Explanation:

5 0

3 years ago

Imagine you are a process safety consultant and you have been tasked to make a metal refinery site DSEAR compliant. What are the

masya89 [10]

Complying with DSEAR involves:

Assessing risks. ...

Preventing or controlling risks. ...

Control measures. ...

Mitigation. ...

Preparing emergency plans and procedures. ...

Providing information, instruction and training for employees. ...

Places where explosive atmospheres may occur ('ATEX' requirements)

hse uk

4 0

2 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

<u>Determine the stress present at the time of fracture for the original plate</u>

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

<u>applying equation 1 determine the value of n for several combinations</u>

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

<em>= 191 mPa </em>

3 0

2 years ago