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3 years ago

A driver of a car traveling 12 m/s applies the breaks. After 2 seconds the speed drops to 8 m/s what is the cars acceleration?

Physics

2 answers:

Afina-wow [57]3 years ago

6 0

Answer:

Deceleration of the car is 2 $\frac{m}{s^{2} }$ .

Explanation:

Given:

initial speed = 12 m/s

final speed = 8 m/s

time = 2 s

To find:

acceleration = ?

Formula used:

According to first kinematic equation:

v = u + a t

Where, v = final speed

u = initial speed

t = time

a = acceleration

Solution:

According to first kinematic equation:

v = u + a t

Where, v = final speed

u = initial speed

t = time

a = acceleration

8 = 12 + 2 a

-4 = 2 a

a = -2 $\frac{m}{s^{2} }$

The negative sign implies that it is deceleration. Thus, deceleration of the car is 2 $\frac{m}{s^{2} }$ .

Minchanka [31]3 years ago

5 0

Answer:

Acceleration value of car = -2 $m/s^2$

Explanation:

We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Here a car is traveling at 12 m/s, so initial velocity = 12 m/s

After 2 seconds the speed drops to 8 m/s, so final velocity = 8 m/s

Time taken = 2 seconds

Substituting

8 = 12 + a * 2

2a = -4

a = -2 $m/s^2$

Acceleration value of car = -2 $m/s^2$

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Aneli [31]

Answer: 0.392 m/s

Explanation:

The Doppler shift equation is:

$f'=\frac{V+V_{o}}{V-V_{s}} f$

Where:

$f=8(10)^{4} Hz$ is the actual frequency of the sound wave

$f'=8.002(10)^{4} Hz$ is the "observed" frequency

$V=1570 m/s$ is the speed of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which are the red blood cells

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$

$V_{s}=\frac{1570 m/s(8.002(10)^{4} Hz-8(10)^{4} Hz)}{8.002(10)^{4} Hz}$

Finally:

$V_{s}=0.392 m/s$

3 0

3 years ago

"There are two types of error that can occur when making measurements: systematic and random error. How you correct the error de

Murljashka [212]

Answer:

Systematic error can be corrected using calibration of the measurement instrument, while random error can be corrected using an average measurement from a set of measurements.

Explanation:

Random errors lead to fluctuations around the true value as a result of difficulty taking measurements, whereas systematic errors lead to predictable and consistent departures from the true value due to problems with the calibration of your equipment.

Systematic error can be corrected, by calibration of the measurement instrument. Calibration is simply a procedure where the result of measurement recorded by an instrument is compared with the measurement result of a standard value.

Random error can be corrected using an average measurement from a set of measurements or by Increasing sample size.

8 0

3 years ago

Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with

Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is $2.728\times 10^{-3}\,\frac{m}{s^{2}}$ ( $\frac{2.784}{10000}\cdot g$ , where $g = 9.8\,\frac{m}{s^{2}}$ ).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration ( $g$ ), in meters per square second, is directly proportional to the mass of the Earth ( $M$ ), in kilograms, and inversely proportional to the distance from the center of the Earth ( $r$ ), in meters:

$g = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$G$ - Gravitational constant, in cubic meters per kilogram-square second.

$M$ - Mass of the Earth, in kilograms.

$r$ - Distance from the center of the Earth, in meters.

If we know that $G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}$ , $M = 5.972\times 10^{24}\,kg$ and $r = 382.26\times 10^{6}\,m$ , then the free-fall acceleration at the orbit of the Moon is: