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2 years ago

Date Page 2. A bicycle of 15kg is moving with the velocity of 10m/s. calculate the kinetic energy.

Physics

1 answer:

QveST [7]2 years ago

3 0

Answer:

$\boxed {\boxed {\sf 750 \ J}}$

Explanation:

Kinetic energy is the energy an object possesses due to motion. It is calculated using the following formula.

$E_k= \frac{1}{2} mv^2$

In this formula, m is the mass and v is the velocity.

The bicycle has a mass of 15 kilograms and a velocity of 10 meters per second.

m= 15 kg
v= 10 m/s

Substitute the values into the formula.

$E_k= \frac {1}{2} (15 \ kg)(10 \ m/s)^2$

Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s = 100 m²/s²

$E_k= \frac{1}{2} (15 \ kg)( 100 \ m^2/s^2)$

Multiply all the values together.

$E_k= \frac{1}{2} ( 1500 \ kg*m^2/s^2)$

$E_k= 750 \ kg*m^2/s^2$

1 kilogram meter squared per second squared is equal to 1 Joule. Therefore, our answer of 750 kg*m²/s² is equal to 750 J.

$E_k= { 750 \ J$

The kinetic energy of the bicycle is 750 Joules.

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Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

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Date Page 2. A bicycle of 15kg is moving with the velocity of 10m/s. calculate the kinetic energy. ​

Date Page 2. A bicycle of 15kg is moving with the velocity of 10m/s. calculate the kinetic energy.