Accidents occur as a result of ____ and ___

A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h i

dmitriy555 [2]

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

${v}^{2} - {u}^{2} = 2as \\ {20}^{2} - {10}^{2} = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a = \frac{300}{200 } = \frac{3}{2} \: m {s}^{ - 2}$

Now, we substitute v, u, and a in the formula

$v = u + at \\ 20 = 10 + \frac{3}{2} t \\ \frac{3}{2} t = 10 \\ 3t = 20 \\ t = \frac{20}{3} = 6.67 \: seconds$

Please mark Brainliest if this helps!

6 0

3 years ago

A 50 (ohm) lossless transmission line is terminated in a load with impedance Z= (30-j50) ohm. the wavelength is 8cm. Determine:

Serjik [45]

Answer:

Reflection Coefficient = $0.57e^{-i79.8}$

SWR=3.65

Position of $V_{max} =3.11cm$

position of $i_{max} =1.11cm$

Explanation:

To determine the above answers, let outline the useful formulas

refection coefficient $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$ .

where terminal impednce = (30-i50)Ω

characteristics impedance= 50Ω

Secondly, the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

Now let us substitute values and solve,

a. $p=\frac{terminalimpednce -characteristics impedance }{terminalimpednce +characteristics impedance } \\$

$p=\frac{(30-i50)-50}{(30-i50)-50} \\$

$p=\frac{-20-i50}{80-i50} \\$

multiplying the numerator and denominator by the conjugate of the denominator. we have

$p=\frac{-20-i50}{80-i50}*\frac{80+i50}{80+i50}\\$

by carrying out careful operation, we arrived at

$p=\frac{900-i5000}{8900} \\p=0.1011-i0.56179\\$

To express in polar form i.e $re^{i alpha}$

$r=\sqrt{0.1011^{2}+0.56179^{2}} \\r=0.57\\$

to get the angle

alpha= $tan^{-1} \frac{0.56179}{0.1011} \\alpha=-79.8\\$

hence the Reflection Coefficient,<em>p</em> = $0.57e^{-i79.8}$

b. we now determine the Standing Wave Ratio, $SWR=\frac{1+/p/}{1-/p/}$

$swr=\frac{1+0.57}{1-0.57} =3.65\\$

c. to determine the position of the maximum voltage nearest to the load,

we use the equation

$Position of V_{max}=\frac{\alpha λ}{4\pi}+\frac{λ}{2}\\$

were $λ$ is the wavelength of 8cm

lets convert α to rad by multiplying by π/180

$Position of V_{max}=\frac{-79.8 *8cm*\pi}{4\pi*180 } +\frac{8cm}{2} \\$

$Position of V_{max}=-0.89+4.0=3.11cm\\$ .

d. also were we have minimum voltage,there the maximum current will exist, to find this position nearest to the load

$Position of v_{min}=Position of V_{max}-\frac{λ}{4} \\\\Position of v_{min}=3.11cm-\frac{8cm}{4}=1.11cm$ .

since the voltage minimum occure at 1.11cm. we can conclude that the current maximum also occur at this point i.e 1.11cm

3 0

3 years ago

Make a proposal to add a small pizza shop to a historical part of town. How could it be designed to “fit” into the area?

spin [16.1K]

It could be designed because if we had a small little pizza shop it would save our problems of going through all the trouble just for pizza

7 0

3 years ago

Read 2 more answers

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

Multiple Choice

mote1985 [20]

I think it’s manufacturing

7 0

3 years ago

Accidents occur as a result of ____ and ____.

Accidents occur as a result of and .