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Alisiya [41]
2 years ago
6

The rear wheel of a 150 kg tricycle accidentally ran over a nail. The circular tip of the nail has a diameter of 0.001 m. How mu

ch pressure did the wheel of the tricycle experience assuming that the entire weight of the tricycle is concentrated on the rear wheel that ran over the nail.
Physics
1 answer:
Brut [27]2 years ago
8 0

The pressure experienced by the wheel of the tricycle is 1.85×10¹⁰ N/m²

<h3>What is pressure?</h3>

Pressure can be defined as the force acting perpendicular per unit area on the surface of a body. The S.I unit of pressure is N/m²

To calculate the pressure of the wheel of the tricycle, we use the formula below

Formula:

  • P = 4mg/πd².............. Equation 1

Where:

  • P = Pressure of the tricycle wheel
  • m = mass of the wheel
  • d = diameter of the wheel
  • g = acceleration due to gravity
  • π = pie.

From the question,

Given:

  • m = 150 kg
  • d = 0.001 m
  • g = 9.8 m/s²
  • π = 3.14

Substitute these values into equation 1

  • P = 4(150)(9.8)/(0.001²×3.14)
  • P = 5880/0.00000314
  • P = 1.85×10¹⁰ N/m²

Hence, The pressure experienced by the wheel of the tricycle is 1.85×10¹⁰ N/m².
Learn more about pressure here: brainly.com/question/25736513

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Hot coffee in a mug cools over time and the mug warms up. Which describes the energy in this system?
givi [52]
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3 0
3 years ago
How much work must be done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners
Natasha_Volkova [10]

Answer:

Potential\ Energy=Work \ Done=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

V is the potential difference

Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

k is Coulomb Constant=8.99*10^9 Nm^2/C^2

q is the charge on electron=-1.6*10^-19 C

r is the distance=3.0*10^{-10}m

For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

Potential\ Energy=3*\frac{(8.99*10^9)(-1.6*10^{-19})^2}{3*10^{-10}}\\Potential\ Energy=2.301*10^{-18} J

Work  done to bring three electrons from a great distance apart to 3.0×10−10 m from one another (at the corners of an equilateral triangle) is 2.301*10^{-18} Joules

7 0
3 years ago
An electric field of 280000 N/C points due west at a certain spot. What is the magnitude of the force that acts on a charge of -
nordsb [41]
<h2>The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.</h2>

Explanation:

Electric field is the ratio of force and charge.

Electric field, E = 280000 N/C

Charge, q = -7.9 C

We have

                 E=\frac{F}{q}\\\\280000=\frac{F}{7.9}\\\\F=280000\times 7.9\\\\F=2.21\times 10^6N

The magnitude of the force that acts on a charge of -7.9C at this spot is 2.21 x 10⁶ N.

4 0
4 years ago
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