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Sholpan [36]
3 years ago
11

A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Det

ermine the average acceleration of the car during this time interval.
Physics
1 answer:
FromTheMoon [43]3 years ago
3 0

Answer:

a=2.5\ m/s^2

Explanation:

Given that,

Initial speed of the car, u = 14 m/s

Finally, it comes to rest, v = 0

Time, t = 5.6 s

We need to find the average acceleration of the car during this time interval. We know that,

a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2

So, the acceleration of the car is 2.5\ m/s^2 in the opposite direction of motion.

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In order to determine the acceleration of the block, use the following formula:

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Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

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by solving for a, you obtain:

a=\frac{kx}{m}

In this case, you have:

k: spring constant = 100N/m

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Replace the previous values of the parameters into the expression for a:

a=\frac{(\frac{100N}{m})(0.02m)}{0.2\operatorname{kg}}=10\frac{m}{s^2}

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The charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Similar charges repel each other, whereas charges that are opposed attract each other.

Given data;

Electric force,F = 9 × 10 ⁻⁹ N

Distance between charges,d = 7 × 10⁻⁴ m

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From Columb's law;

\rm F = K \frac{q_1q_2}{d^2} \\\\ 9 \times 10^{-9}  = 9 \times 10^9 \frac{q^2}{(7 \times 10^{-4})^2} \\\\ q^2 = 4.9 \times 10^{-25} \\\\  q = 7 \times 10^{-13} \ C

Hence the charge on each of the equally charged drops of hairspray willl be 7 × 10 ⁻¹³ C

To learn more about Columb's law refer to the link;

brainly.com/question/1616890

#SPJ1

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