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Dmitry [639]
3 years ago
11

A car with an initial speed of 4.30 m/s accelerates

Physics
1 answer:
Ivenika [448]3 years ago
3 0

Heya!!

For calculate final velocity, lets applicate formula

                                                \boxed{V=V_o+a*t}

                                                 <u>Δ   Being   Δ</u>

                                            V = Final Velocity = ?

                                         Vo = Initial velocity = 4,3 m/s

                                          a = Aceleration = 3 m/s²

                                                 t = Time =5 s

⇒ Let's replace according the formula:

\boxed{V=4,3\ m/s +3\ m/s* 5\ s}

⇒ Resolving

\boxed{V=19,3\ m/s}

Result:

The velocity after 5 sec is <u>19,3 meters per second (m/s)</u>

For calculate distance, lets applicate formula

                                              \boxed{x=V_o*t+\dfrac{a*t^{2}}{2}}}

                                                 <u>Δ   Being   Δ</u>

                                               x = Distance = ?

                                         Vo = Initial velocity = 4,3 m/s

                                          a = Aceleration = 3 m/s²

                                                 t = Time =5 s

⇒ Let's replace according the formula:

\boxed{x=4,3 * 5 +\dfrac{3*5^{2}}{2}}}

⇒ Resolving

\boxed{x = 59 \ m}

Result:

The distance is <u>59 meters.</u>

Good Luck!!

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Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a −80-nC charge at x = −4 m on
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Answer:

V = 48 Volts

Explanation:

Since we know that electric potential is a scalar quantity

So here total potential of a point is sum of potential due to each charge

It is given as

V = V_1 + V_2 + V_3

here we have potential due to 50 nC placed at y = 6 m

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Now potential due to -80 nC charge placed at x = -4

V_2 = \frac{kQ}{r}

V_2 = \frac{(9\times 10^9)(-80 \times 10^{-9})}{12}

V_2 = -60 Volts

Now potential due to 70 nC placed at y = -6 m

V_3 = \frac{kQ}{r}

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3 0
3 years ago
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Newton's third law, if i'm not mistaken.

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18. Direction.

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20. Speed.

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Because the velocity of an object depends on direction as well as speed, the velocity of an object can change even as the speed of the object remains constant. Thus, the velocity would change only when the direction or speed of the object changes.

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2 years ago
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