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3 years ago

Please help me solve this question ><

Physics

1 answer:

Snowcat [4.5K]3 years ago

4 0

Option B, 1.O m/s is correct

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Define Archemedics principle?

azamat

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.

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3 years ago

A square conducting plate 52.0 cm on a side and with no net charge is placed in a region, where there is a uniform electric fiel

Mrac [35]

Explanation:

Given: uniform electric field E= 82.0 kN/C.

a) charge density σ =ε_0 E.

therefore, $\sigma =82\times10^3\times3.85\times10^{-12}\\=0.0000003157= 315.7 nC/m^2$

b)Total charge on each face = σA

q=σA

$=315.7\times10^{-9}\times52\times10^{-4}\\=1.614\times10^{-9}= 1.614 \text{ nC}\\\text{Similarly on the other face } = -1.614 \text{ nC}$

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4 years ago

What feature on the periodic table separates the metals and non metals?

sertanlavr [38]

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