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1 year ago

Define friction State the law of frictionGive two factors that may affect friction

Physics

1 answer:

umka21 [38]1 year ago

3 0

Friction:

Friction is the resistive force that acts on the body in a direction opposite to the direction of motion of the same object.

Laws of friction:

1. The frictional force acting on the object is proportional to the normal force acting on the object.

For example, if the object of mass m is moving along the horizontal surface, the normal force acting on the object is mg.

Thus, the frictional force acting on the object is,

$\begin{gathered} F_r\propto F_N \\ F_r=\mu F_N \end{gathered}$

where F_N is the normal force acting on the object,

$\mu\text{ is the coefficient of friction}$

The diagrammatic representation of the frictional force acting on the object is,

2. The frictional force acting on the object depends upon the nature of the surface in contact with the object.

3. The frictional force does not depend upon the area of contact of an object with the surface.

4. The kinetic friction acting on the object is independent of the velocity of the object.

5. The coefficient of static friction is more than the coefficient of kinetic friction.

Two factors on which the frictional force depends are:

1. The frictional force depends on the adhesion between the two bodies in contact.

2. The frictional force acting on the depends upon the roughness of the surface on which the object is moving.

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Light is incident on an air-water interface at an angle of 30 degree to the normal. What angle does the refracted ray make with

Vikentia [17]

Answer:

22 degree

Explanation:

Angle of incidence, i = 30 degree

the refractive index of water with respect to air is 4/3.

As the ray of light travels from rarer medium to denser medium, that mean air to water, the refraction takes place.

According to Snell's law,

Refractive index of water with respect to air = Sin i / Sin r

Where, r be the angle of refraction

4 / 3 = Sin 30 / Sin r

0.75 = 2 Sin r

Sin r = 0.375

r = 22 degree

Thus, the angle of refraction is 22 degree.

6 0

3 years ago

A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

The initial height of the water in a sealed container of diameter 100.0 cm is 5.00 m. The air pressure inside the container is 0

katen-ka-za [31]

Answer:

a) F = 2.66 10⁴ N, b) h = 1.55 m

Explanation:

For this fluid exercise we use that the pressure at the tap point is

Exterior

P₂ = P₀ = 1.01 105 Pa

inside

P₁ = P₀ + ρ g h

the liquid is water with a density of ρ=1000 km / m³

P₁ = 0.85 1.01 10⁵ + 1000 9.8 5

P₁ = 85850 + 49000

P₁ = 1.3485 10⁵ Pa

the net force is

ΔP = P₁- P₂

Δp = 1.3485 10⁵ - 1.01 10⁵

ΔP = 3.385 10⁴ Pa

Let's use the definition of pressure

P = Fe / A

F = P A

the area of a circle is

A = pi r² = [i d ^ 2/4

let's reduce the units to the SI system

d = 100 cm (1 m / 100 cm) = 1 m

F = 3.385 104 pi / 4 (1) ²

F = 2.66 10⁴ N

b) the height for which the pressures are in equilibrium is

P₁ = P₂

0.85 P₀ + ρ g h = P₀

h = $\frac{P_o ( 1-0.850)}{\rho \ g}$

h = $\frac{1.01 \ 10^5 ( 1 -0.85)}{1000 \ 9.8}$

h = 1.55 m

4 0

3 years ago

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally h

zysi [14]

Complete question:

What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)

Answer:

The peak emf generated by the coil is 15.721 kV

Explanation:

Given;

Radius of coil, r = 0.250 m

Number of turns, N = 500-turn

time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s

magnetic field strength, B = 0.425 T

Induced peak emf = NABω

where;

A is the area of the coil

A = πr²

ω is angular velocity

ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s

Induced peak emf = NABω

= 500 x (π x 0.25²) x 0.425 x 376.738

= 15721.16 V

= 15.721 kV

Therefore, the peak emf generated by the coil is 15.721 kV

5 0

3 years ago

What is the momentum of a 1, 500 kg car traveling at 3m/s?

ruslelena [56]

4,500 kgm/s bc 1,500x3

6 0

3 years ago

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