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2 years ago

While traveling along a highway, a driver slows from 27 m/s to 13 m/s in 11 seconds. What is the automobile’s acceleration?

Physics

1 answer:

Travka [436]2 years ago

4 0

Answer:

$a=-1.27\ m/s^2$

Explanation:

Given that,

The initial speed of a driver, u = 27 m/s

Final speed of the driver, v = 13 m/s

Time, t = 11 s

We need to find the automobile's acceleration. We know that the acceleration of an object is equal to the change in speed divided by time taken.

$a=\dfrac{13-27}{11}\\\\a=-1.27\ m/s^2$

So, the magnitude of automobile's acceleration is $1.27\ m/s^2$ .

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The graph below shows the position of an ant as it crawls over a flat picnic blanket. The total time for the ant to go from the

Mice21 [21]

The average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

The correct answer is option D.

In the given graph, we can deduce the following;

the total time of the motion, = 1 mins + 45 s = 60 s + 45 s = 105 s

The average speed of the ant is calculated as;

$average \ speed = \frac{total \ distance }{total \ time }$

The total distance from the graph is calculated as follows;

first horizontal distance from 2 cm to 8 cm = 8 - 2 = 6 cm

first upward distance from 3 cm to 5 cm = 5 - 3 = 2 cm

second horizontal distance from 8 cm to 6 cm = 8 - 6 = 2 cm

second upward distance from 5 cm to 12 cm = 12 - 5 = 7 cm

third horizontal distance from 6 cm to 13 cm = 13 - 6 = 7 cm

fourth downward distance from 12 cm to 9 cm = 3 cm

final horizontal distance from 13 cm to 15 cm = 2cm

The total distance = (6 + 2 + 2 + 7 + 7 + 3 + 2) cm = 29 cm

$average \ speed = \frac{total \ distance }{total \ time } = \frac{29 \ cm}{105 \ s} = 0.276 \ cm/s$

The average velocity is calculated as the change in displacement per change in time.

The displacement is the shortest distance between the start and end positions.

This shortest distance is the straight line connecting the start and end position. Call this line P

From the end position at x = 15 cm, draw a vertical line from y = 9 cm, to y = 3 cm. The displacement = 9 cm - 3 cm = 6 cm

Also, draw a horizontal line from start at x = 2 cm to x = 15 cm. The displacement = 15 cm - 2 cm = 13 cm

Notice, you have a right triangle, now calculate the length of line P.

↓end

↓

↓ 6cm

↓

start -------------13 cm------------

Use Pythagoras theorem to solve for P.

$P^2 = 6^2 + 13^2\\\\P^2 = 36 + 169\\\\P^2 = 205\\\\P= \sqrt{205} \\\\P = 14.318 \ cm$

The average velocity of the ant is calculated as;

$average \ velocity= \frac{\Delta displacemnt }{total\ time }= \frac{14.318 \ cm}{105 \ s} = 0.136 \ cm/s \\\\$

Thus, the average speed of the ant is 0.276 cm/s and the average velocity is 0.136 cm/s.

Learn more here: brainly.com/question/589950

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2 years ago

Why a raw egg shell disolves when submerge in vineger

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If you soak this egg shell in vinegar (which is about 4% acetic acid), you start a chemical reaction that dissolves the calcium carbonate shell. The acetic acid reacts with the calcium carbonate in the egg shell and releases carbon dioxide gas that you see as bubbles on the shell.

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Answer and Explanation:

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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

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$dq = \frac{3Qr^2dr}{R^3}$

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$E = \int dW$

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