2 years ago

What should you release to re-establish vehicle control and tire traction?

Engineering

1 answer:

MrMuchimi2 years ago

6 0

Answer: The accelerator and the brakes.

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11. As __and___ prices continued to rise in the late 1960’s and 70's, 4 and 6 cylinder engines began to make a comeback.

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Answer:

Explanation:

whats the answeres two the question do they have a choise for you

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6 months ago

An automobile having a mass of 1100 kg initially moves along a level highway at 110 km/h relative to the highway. It then climbs

Greeley [361]

Answer:

Change in kinetic energy=-513.652 KJ

Change in potential energy=431.64KJ

Explanation:

We are given that

Mass of an automobile , m=1100 kg

Initial speed, u=110 km/h= $110\times \frac{5}{18}=30.56 m/s$

Where $1 km/h=5/18 m/s$

Height , $h_2=40 m$

$g=9.81 m/s^2$

Final speed, v=0

Change in kinetic energy, $\Delta K.E=\frac{1}{2}m(v^2-u^2)$

$\Delta K.E=\frac{1}{2}(1100)(0-(30.56)^2)=-513652.48 J$

$\Delta K.E=-\frac{513652.48}{1000}=-513.652 KJ$

Where 1 KJ=1000 J

Change in potential energy, $\Delta P.E=mgh(h_2-h_1)$

Initially height, h1=0

Using the formula

$\Delta P.E=1100\times 9.81(40-0)$

$\Delta P.E=431640J$

$\Delta P.E=431.64KJ$

6 0

2 years ago

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea

vovangra [49]

Answer: 33.35 minutes

Explanation:

A(t) = A(o) *(.5)^[t/(t1/2)]....equ1

Where

A(t) = geiger count after time t = 100

A(o) = initial geiger count = 400

(t1/2) = the half life of decay

t = time between geiger count = 66.7 minutes

Sub into equ 1

100=400(.5)^[66.7/(t1/2)

Equ becomes

.25= (.5)^[66.7/(t1/2)]

Take log of both sides

Log 0.25 = [66.7/(t1/2)] * log 0.5

66.7/(t1/2) = 2

(t1/2) = (66.7/2 ) = 33.35 minutes

3 0

2 years ago

Pay attention to the following questions!

kupik [55]

Answer: the correct answer is D

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1 year ago

A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L

zzz [600]

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD = 30 mg/L × dilution factor

Original BOD = 30 mg/L × 10 = 300 mg/L

$L_o = \frac{BOD}{1-e^{-5t}}$

here $L_o$ is the ultimate BOD ; BOD is the biochemical oxygen demand ; t = 0.20 /day

$L_o = \frac{300}{1-e^{-5(0.20)}}$

$L_o = 474.59 \ mg/L$

3 0

3 years ago