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Artemon [7]
2 years ago
7

A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

the cyclist’s initial velocity?.
Physics
1 answer:
mel-nik [20]2 years ago
6 0

Answer:

\boxed {\boxed {\sf v_i= 4 \ m/s}}

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

v_f= v_i + at

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

  • v_f= 16 m/s
  • a= 1.2 m/s²
  • t= 10 s

Substitute the values into the formula.

16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)

Multiply.

16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)

16 \ m/s = v_i + 12 \ m/s

We are solving for the initial velocity, so we must isolate the variable v_i. Subtract 12 meters per second from both sides of the equation.

16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s

4 \ m/s = v_i

The cyclist's initial velocity is <u>4 meters per second.</u>

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A. - 0.017N. It acts to the left.

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Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

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Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

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B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

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C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

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F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

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