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3 years ago

A hammer has a mass of 1 kg. What is its weight (i) on Earth (ii) on the

Physics

1 answer:

solong [7]3 years ago

4 0

Given mass= 1kg

Weight on earth = mg(gravity of earth) = 9.8N

weight on moon = mg(gravity of moon)= 1.62N

weight on outer space mg(gravity outer space = 0) = 0N

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An automobile follows a circular road whose radius is 50 m. Let x and y respectively denote the eastern and northern directions,

IceJOKER [234]

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration ( $a$ ) can be defined by this ordinary differential equation in terms of distance:

$a = v\cdot \frac{dv}{ds}$ (1)

Where:

$v$ - Speed of the automobile, measured in meters per second.

$s$ - Distance travelled by the automobile, measured in meters.

If we know that $v = 0.5\cdot s - 0.0025\cdot s^{2}$ , then the equation of acceleration is:

$a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)$

$a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)$

$a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)$

But distance covered by the vehicle is defined by the following formula:

$s = \theta \cdot r$ (2)

Where:

$\theta$ - Arc angle, measured in radians.

$r$ - Radius, measured in radians.

Then, we expand (1) by means of this result:

$a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)$

$a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)$

$a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r$

And finally we get the following third order polynomial:

$1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0$ (3)

If we know that $r = 50\,m$ , $a = 0.6\cdot g$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the polynomial becomes into this:

$1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0$ (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

$\theta_{1} \approx 4.365\,rad$ , $\theta_{2} \approx 0.818+i\,0.441\,rad$ , $\theta_{3}\approx 0.818 -i\,0.441\,rad$ , $\theta_{4} \approx 1.563\,rad$

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid <em>first</em>. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: ( $r = 50\,m$ , $\theta \approx 1.563\,rad$ )