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Maurinko [17]
2 years ago
14

What is the last item on the vehicle starting checklist

Engineering
1 answer:
larisa86 [58]2 years ago
4 0

The last item on the vehicle starting checklist before a driver begins to travel is for the driver to release the service brake.

<h3>What does it mean to release the brake?</h3>

The Release Parking Brake message implies that your parking brake is said to be currently engaged and as such one needs to  release the parking brake before one can be able to drive the vehicle.

Therefore, The last item on the vehicle starting checklist before a driver begins to travel is for the driver to release the service brake.

See full question below

what is the last item on the vehicle starting checklist before a driver begins to travel?

Learn more about checklist from

brainly.com/question/16041035

#SPJ11

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The following C program asks the user for two input null-terminated strings, each stored in uninitialized 100-byte buffer, and c
marissa [1.9K]

Answer:

Code is given below:

Explanation:

.data  

str1: .space 20  

str2: .space 20  

msg1:.asciiz "Please enter string (max 20 characters): "  

msg2: .asciiz "\n Please enter string (max 20 chars): "  

msg3:.asciiz "\nSAME"  

msg4:.asciiz "\nNOT SAME"  

.text

.globl main

main:  

   li $v0,4        #loads msg1  

   la $a0,msg1  

   syscall

   li $v0,8

   la $a0,str1

   addi $a1,$zero,20

   syscall          #got string to manipulate

   li $v0,4        #loads msg2

   la $a0,msg2

   syscall

   li $v0,8

   la $a0,str2

   addi $a1,$zero,20

   syscall         #got string  

       la $a0,str1             #pass address of str1  

   la $a1,str2         #pass address of str2  

   jal methodComp      #call methodComp  

   beq $v0,$zero,ok    #check result  

   li $v0,4

   la $a0,msg4

   syscall

   j exit

ok:  

   li $v0,4  

   la $a0,msg3  

   syscall  

exit:  

   li $v0,10  

   syscall  

methodComp:  

   add $t0,$zero,$zero  

   add $t1,$zero,$a0  

   add $t2,$zero,$a1  

loop:  

   lb $t3($t1)         #load a byte from each string  

   lb $t4($t2)  

   beqz $t3,checkt2    #str1 end  

   beqz $t4,missmatch  

   slt $t5,$t3,$t4     #compare two bytes  

   bnez $t5,missmatch  

   addi $t1,$t1,1      #t1 points to the next byte of str1  

   addi $t2,$t2,1  

   j loop  

missmatch:    

   addi $v0,$zero,1  

   j endfunction  

checkt2:  

   bnez $t4,missmatch  

   add $v0,$zero,$zero  

endfunction:  

   jr $ra

3 0
3 years ago
For the same cross-sectional area, which column provides the higher buckling load: a circular bar or a circular tube?
juin [17]

Answer:

Circular tube

Explanation:

Now for better understanding lets take an example

Lets take

Diameter of solid bar= 4\sqrt{2} cm

Outer diameter of tube =6 cm

Inner diameter of tube=2 cm

So from we can say that both tubes have equal cross sectional area.

We know that buckling load is given as P = \dfrac{\pi ^2EI}{L_e^2}      

If area moment of inertia(I) is high then buckling load will be high.

We know that  area moment of inertia(I)

For circular tube I = \dfrac{\pi }{64}(D_o^4-D_i^4)

For circular bar I = \dfrac{\pi }{64}D^4  

Now by putting the values

    For circular tube I=62.83 cm^4

  For circular bar I=50.26 cm^4

So we can say that for same cross sectional area the  area moment of inertia(I) is high for tube as compare to bar.So buckling load  will be higher in tube as compare to bar.

3 0
3 years ago
An existing building is suffering from cracks in the exterior walls. The investigating engineer wants to ensure that the foundat
jek_recluse [69]

Answer:

18 ft^{2}

Explanation:

Soil bearing pressure=\frac {Load}{Area}

Since we're given pressure of 2500 psf and load of 45000 pounds

The area=\frac {45000}{2500}=18

Therefore, the smallest area of safe footings should not be less than 18 ft^{2}

6 0
3 years ago
A student engineer is given a summer job to find the drag force on a new unmaned aerial vehicle that travels at a cruising speed
yan [13]

Answer:

b. 1232.08 km/hr

c. 1.02 kn

Explanation:

a) For dynamic similar conditions, the non-dimensional terms R/ρ V2 L2 and ρVL/ μ should be same for both prototype and its model. For these non-dimensional terms , R is drag force, V is velocity in m/s, μ is dynamic viscosity, ρ is density and L is length parameter.

See attachment for the remaining.

3 0
4 years ago
Identify which sound type each line contains.
nydimaria [60]

Answer:i can not see it

Explanation:

4 0
3 years ago
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