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3 years ago

Which is usually greater for a pair of surfaces, the coefficient of static friction or the coefficient of sliding friction? why

Physics

1 answer:

lorasvet [3.4K]3 years ago

5 0

Explanation :

Static friction is the frictional force between two objects that are at rest. While sliding friction is the frictional force between two objects in contact and are sliding w.r.t each other.

Static friction is usually greater than sliding friction because in static friction the contact forces is more and the interlocking between objects is tight as compared to sliding friction.

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A 210 Ohm resistor uses 9.28 W of

IgorLugansk [536]

Resistance=R=210Ohm
Power=9.28W=P

Current=I

$\boxed{\sf P=I^2R}$

$\\ \sf\longmapsto I^2=\dfrac{P}{R}$

$\\ \sf\longmapsto I^2=\dfrac{9.28}{210}$

$\\ \sf\longmapsto I^2\approx0.04$

$\\ \sf\longmapsto I\approx\sqrt{0.04}$

$\\ \sf\longmapsto I\approx\sqrt{\dfrac{4}{100}}$

$\\ \sf\longmapsto I\approx\dfrac{\sqrt{4}}{\sqrt{100}}$

$\\ \sf\longmapsto I\approx\dfrac{2}{10}$

$\\ \sf\longmapsto I\approx0.2A$

4 0

3 years ago

A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebbl

Leona [35]

<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

<em>✔ We have: KE = PE (potential energy) </em>

The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2

PE1 = PE2 ⇔ PE1/PE2 = 1

$\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\ \frac{m_1}{m_2\times 4} = 1 \\ \\ \frac{m_1}{m_2} = 4$

The mass m1 is therefore 4 times greater than that of the stone of mass m2.

8 0

2 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$