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3 years ago

a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration

Physics

1 answer:

Aneli [31]3 years ago

8 0

Answer:

Acceleration is 2.78 m/s²

Explanation:

Since the car is from rest, initial velocity, u is zero.

from first newton's equation of motion:

${ \bf{v = u + at}} \\$

v is final velocity

u is initial velocity

a is acceleration

t is time

70 km/h is (70 × 1000 ÷ 3600) m/s = 19.44 m/s

${ \sf{19.44 = 0 + (a \times 7)}} \\ { \sf{7a = 19.44 }} \\ { \sf{a = 2.78 \: {ms}^{ - 2} }}$

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The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

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a 0.40kg soccer ball approaches a player horizontally with a velocity of 18m/s north. the player strikes the ball and causes it

denis23 [38]

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a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration​

a car advertisement States that a certain car can accelerate from rest to70km/h in 7seconds find the car's average acceleration