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3 years ago

If you detect that something’s wrong while lowering a lift, you should try to get the vehicle down as fast as possible

1 answer:

4 0

False, it depends on the situation. If the lift is tilting or anything like I would then get down. Certain training will say to get out and see if you can keep lowering,

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The thrust angle is checked by referencing

anygoal [31]

In Engineering, the thrust angle is checked by referencing: C. vehicle centerline.

<h3>What is a thrust angle?</h3>

A thrust angle can be defined as an imaginary line which is drawn perpendicularly from the centerline of the rear axle of a vehicle, down the centerline.

This ultimately implies that, the thrust angle is a reference to the centerline (wheelbase) of a vehicle, and it confirms that the two wheels on both sides are properly angled within specification.

Read more on thrust angle here: brainly.com/question/13000914

#SPJ1

5 0

2 years ago

The coefficient of static friction for both wedge surfaces is μw=0.4 and that between the 27-kg concrete block and the β=20° inc

balandron [24]

Assuming the wedge has an angle of 5°.The minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

<h3>Minimum value of force P</h3>

First step

Using this formula to find the weight of the block

W=mg

W=27×9.81

W=264.87 N

Second step

Angles of friction ∅A and ∅B

∅A=tan^-1(μA)

∅A=tan^-1(0.70)

∅A=34.99°

∅B=tan^-1(μB)

∅B=tan^-1(0.40)

∅B=21.80°

Third step

Equate the sum of forces in m-direction to 0 in order to find the reaction force at B.

∑fm=0

W sin (∅A+20°) + RB cos (∅B+∅A)=0

264.87 sin(34.99°+20°) + RB cos (21.80°+34.99°)=0

216.94+0.5477Rb=0

RB=216.94/0.5477

RB=396.09 N

Fourth step

Equate the sum of forces in x-direction to 0 in order to find force Rc.

∑fx=0

RB cos (∅B) - RC cos (∅B+ 5°)=0

396.09 cos(21.80°) - RC cos (21.80°+5°)=0

RC=396.09 cos(21.80°)/cos(26.80°)

RC=412.02 N

Last step

Equate the sum of forces in y-direction to 0 in order to find force P required to move the block up the incline.

∑fy=0

RB sin (∅B) + RC sin (∅B)-P=0

P=Rb sin (∅B) + RC sin (5°+∅B)

P=396.09 sin(21.80°) +412.02sin (5°+21.80°)

P=322.84 N

Inconclusion the minimum value of the force P that is required to begin moving the block up the incline is: 322.84 N.

Learn more about Minimum value of force P here:brainly.com/question/20522149

7 0

2 years ago

A heat pump with an ideal compressor operates between 0.2 MPa and 1 MPa. Refrigerant R134a flows through the system at a rate of

solmaris [256]

Answer:

The mass flow rate of refrigerant is 0.352 kg/s

Explanation:

Considering the cycle of an ideal heat pump, provided in the attachment, we first find enthalpy at state B and D. For that purpose, we use property tables of refrigerant R134a:

<u>At State A</u>:

From table, we see the enthalpy and entropy value of saturated vapor at 0.2 MPa. Therefore:

ha = 244.5 KJ/kg

Sa = 0.93788 KJ/kg.k

<u>At State B</u>:

Since, the process from state A to B is isentropic. Therefore,

Sb = Sa = 0.93788 KJ/Kg

From table, we see the enthalpy value of super heated vapor at 1 MPa and Sb. Therefore:

hb = 256.85 KJ/kg (By interpolation)

<u>At State C</u>:

From table, we see the enthalpy and entropy value of saturated liquid at 1 MPa. Therefore:

hc = 107.34 KJ/kg

Now, from the diagram it is very clear that:

Heat Loss = m(hb = hc)

m = (Heat Loss)/(hb - hc)

where,

m = mass flow rate = ?

Heat Loss = (180,000 Btu/hr)(1.05506 KJ/1 Btu)(1 hr/3600 sec)

Heat Loss = 52.753 KW

Therefore,

m = (52.753 KJ/s)/(256.85 KJ/kg - 107.34 KJ/kg)

5 0

3 years ago

A standard penetration test has been conducted on a coarse sand at a depth of 16 ft below the ground surface. The blow counts ob

scoray [572]

Solution :

Given :

The number of blows is given as :

0 - 6 inch = 4 blows

6 - 12 inch = 6 blows

12 - 18 inch = 6 blows

The vertical effective stress $=1500 \ lb/ft^2$

$= 71.82 \ kN/m^2$

$\sim 72 \ kN/m^2$

Now,

$N_1=N_0 \left(\frac{350}{\bar{\sigma}+70} \right)$

$N_1 =$ corrected N - value of overburden

$\bar{\sigma}=$ effective stress at level of test

0 - 6 inch, $N_1=4 \left(\frac{350}{72+70} \right)$

= 9.86

6 - 12 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

12 - 18 inch, $N_1=6 \left(\frac{350}{72+70} \right)$

= 14.8

$N_{avg}=\frac{9.86+14.8+14.8}{3}$

= 13.14

= 13

8 0

3 years ago

When the psychologist simply records the relationship between two variables...

Wewaii [24]

When a psychologist simply records the relationship between two variables without manipulating them, it is called a correlational study.

The observed relationship does not by itself reveal which variable causes the other. This is the directionally problem. Also, the relationship may be due to a third variable controlling both of the observed variables.

8 0

3 years ago