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-Dominant- [34]

3 years ago

10

A car starts from rest and accelerates along a straight line path in one minute. It finally attains a velocity of 40 meters/seco

nd. What is the car's average acceleration?

1 answer:

romanna [79]3 years ago

7 0

Answer:

it's A

Explanation:

I JUST A FROM THE TOPIC

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Verify that for values of n less than 8, the system goes to a stable equilibrium, but as n passes 8, the equilibrium point becom

irakobra [83]

Answer:

Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.

Explanation:

5 0

2 years ago

What is the base number for the metric system of units? In other words, by what number do we multiply to move up in scale ? ____

Andru [333]

Answer:

10

Explanation:

Each time you want to move up the scale, the number is multiplied by 10. The opposite is true when moving down; you must divide by 10.

6 0

3 years ago

Read 2 more answers

Sound waves travel in air at 343 m/s. The lowest frequency one can hear is 25.0 Hz; the highest frequency is 25.0 kHz. Find the

Slav-nsk [51]

Answer:

13.72 m and 0.01372 m respectively

Explanation:

Wavelength: This can be defined as the distance covered in one complete oscillation. The S.I unit of wavelength is meter (m).

The formula for the speed of a wave is given as

v = λf ............................. Equation 1

Where v = speed of the sound wave, λ = wavelength, f = frequency of the sound wave.

make λ the subject of the equation,

λ = v/f ......................... Equation 2

For the lowest frequency,

Given: f = 25 Hz, v = 343 m/s.

Substitute into equation 2

λ = 343/25

λ = 13.72 m.

For the highest frequency,

Given: f = 25 kHz = 25000 Hz, v = 343 m/s

Substitute into equation 2

λ = 343/25000

λ = 0.01372 m.

The wavelength of sound for 25 Hz and 25 kHz = 13.72 m and 0.01372 m respectively

8 0

3 years ago

Josh and jake are both helping to build a brick wall which is 6 meters in height. They each lay 250 bricks, but josh finishes th

Yakvenalex [24]

Each laid 250 bricks but while Jake was
still working, Josh was lounging in the
shade. Josh has more power but that
power was only on for 3 hours out of 4.5.
Obviously Josh could get more done is less
time as long as he keeps working. Jake will
get the hang of it soon.

6 0

2 years ago

Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

3 years ago