Disorder in the universe increases because

The change in position of an object per unit of time is referred to as

Ivanshal [37]

I think it's speed. Distance/time=speed

4 0

2 years ago

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: ( $r = 50\,m$ , $\theta \approx 1.563\,rad$ )

$s = (1.563\,rad)\cdot (50\,m)$

$s = 78.15\,m$

Lastly, the speed of the automobile at this location is: ( $s = 78.15\,m$ )

$v = 0.5\cdot s - 0.0025\cdot s^{2}$ (4)

$v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}$

$v = 23.806\,\frac{m}{s}$

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

7 0

2 years ago

An ice cube with a mass of 0.0600 kg is placed at the midpoint of a 1.00-m-long wooden board that is propped up at a 49° angle.

Bess [88]

Answer:

Explanation:

distance covered by ice cube = 0.5 m

force acting downwards = mgsin49

force of friction acting along surface upwards

= mgcos49 x .16 ( reaction force of inclined surface R = mgcos49.)

net force acting downwards

mgsin49 - mgcos49 x .16

mg (sin49 - cos49 x .16 )

net acceleration downwards

g (sin49 - cos49 x .16 )

= (.7547 - .105 ) x 9.8

= 6.367 m / s²

s = ut + 1/2 at²

0.5 = 0 + 1/2 x 6.367 x t²

t² = .1570

t = .4 s

b )

Let required inclination be θ

in the second case net acceleration

a = g ( sin θ - .385 cosθ )

s = ut + 1/2 at²

0.5 = 0 + 1/2 x g ( sin θ - .385 cosθ ) x .4 x .4

( sin θ - .385 cosθ ) = .6377

sin θ = .385 cosθ + .6377

θ = 58 degree.

3 0

2 years ago

Which of the following is a constant force?

Kaylis [27]

Answer:

Gravitational force

Explanation:

gravity is what keeps us down on earth so we don`t float away into space and it is constant it does it 24/7 365 days a year and the remaining time earth continues to survive

4 0

3 years ago

Read 2 more answers

A heat engine operates between 200 K and 100 K. In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold rese

Elena-2011 [213]

Answer:

It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J

But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law

6 0

3 years ago