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3 years ago

15

The total weight of a car with passengers is 30,000N. The area of contact of each of the four tyres with the ground is 0.025m2.

Determine the minimum car tyre pressure

1 answer:

vlabodo [156]3 years ago

7 0

Answer:

<em>The minimum pressure car tire pressure is 300,000 Pa</em>

Explanation:

<u>Pressure</u>

It's a measure of the force exerted by an object per unit surface:

$\displaystyle P=\frac{F}{A}$

Where F is the force applied on a surface area A.

We are given the total weight of a car with passengers as Ft=30,000 N and we also know the area of contact of each tire with the ground as $A=0.025 m^2$ .

The minimum pressure each tire can have is when the total weight is evenly distributed in all four tires, thus F=Ft/4=7,500 N.

Calculating the pressure:

$\displaystyle P=\frac{7,500}{0.025}$

P=300,000 Pa

The minimum pressure car tire pressure is 300,000 Pa

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(a) The average speed from A to B would be 1.76 metre per second and the average velocity from A to B would also be 1.76 metre per second

<span>(b) The average speed from A to C would be 1.73 metre per second and the average velocity from A to C would be 0.87 metre per second</span>

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3 years ago

A 2.93 kg particle has a velocity of (2.98 i hat - 3.98 j) m/s.

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Answer:

a) The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

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Explanation:

a) The vectorial equation of momentum is represented by the following expression:

$\vec p = m\cdot \vec v$ (1)

Where:

$\vec p$ - Vector momentum, measured in kilogram-meters per second.

$m$ - Mass of the particle, measured in kilograms.

$\vec v$ - Vector velocity, measured in meters per second.

If we know that $m = 2.93\,kg$ and $\vec v = 2.98\,\hat{i}-3.98\,\hat{j}\,\,\,\left[\frac{m}{s} \right]$ , then the momentum is:

$\vec p = (2.93)\cdot (2.98\,\hat{i}-3.98\,\hat{j})\,\,\,\left[\frac{kg\cdot m}{s} \right]$

$\vec p = 8.731\,\hat{i}-11.661\,\hat{j}\,\,\,\left[\frac{kg\cdot m}{s} \right]$

The x and y components of the momentum are $8.731\,\frac{kg\cdot m}{s}$ and $-11.661\,\frac{kg\cdot m}{s}$ , respectively.

b) The magnitude and direction of momentum are represented by the following expressions:

$\|\vec p \| = \sqrt{p_{x}^{2}+p_{y}^{2}}$ (2)

$\theta = \tan^{-1}\left(\frac{p_{y}}{p_{x}} \right)$ (3)

Where:

$\|\vec p\|$ - Magnitude of momentum, measured in kilogram-meters per second.

$\theta$ - Direction of momentum, measured in sexagesimal degrees.

If we know that $p_{x} = 8.731\,\frac{kg\cdot m}{s}$ and $p_{y} = -11.661\,\frac{kg\cdot m}{s}$ , then the magnitude and direction of momentum are, respectively:

$\|\vec p\| = \sqrt{\left(8.731\,\frac{kg\cdot m}{s} \right)^{2}+\left(-11.661\,\frac{kg\cdot m}{s} \right)^{2}}$

$\|\vec p\| \approx 14.567\,\frac{kg\cdot m}{s}$

$\theta =\tan^{-1}\left(\frac{-11.661\,\frac{kg\cdot m}{s} }{8.731\,\frac{kg\cdot m}{s} } \right)$

$\theta \approx 306.823^{\circ}$

The magnitude and direction of its momentum are approximately 14.567 kilogram-meters per second and 306.823º.

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3 years ago

A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee

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Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

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Electric potential is given as;

V = E*r

where;

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V = ( kq/r²)*r

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V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

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Answer:

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Answer:

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3 years ago