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3 years ago

Name and explain the various types of friction.

Physics

1 answer:

yawa3891 [41]3 years ago

8 0

Answer:

There are four types of friction: static, sliding, rolling, and fluid friction. Static, sliding, and rolling friction occur between solid surfaces. Static friction is strongest, followed by sliding friction, and then rolling friction, which is weakest. Fluid friction occurs in fluids, which are liquids or gases.

Explanation:

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A long, straight wire carries a current of 8.60

Kay [80]

First of all, we need to calculate the magnetic field generated by the wire at a distance r=3.90 cm=0.039 m, which is given by:

$B=\frac{\mu_0 I}{2 \pi r}$

where I=8.60 A is the current. Substituting numbers in the equation, we find

$B=\frac{(4 \pi \cdot 10^{-7})(8.6 A)}{2 \pi (0.039 m)}=4.4 \cdot 10^{-5} T$

And now we can calculate the force exerted on the electron, which is given by:

$F=qvB$

where $q=1.6 \cdot 10^{-19} C$ is the charge of the electron and $v=5.0 \cdot 10^4 m/s$ is its speed. Substituting data in the formula, we find

$F=(1.6 \cdot 10^{-19}C)(5.00 \cdot 10^4 m/s)(4.4 \cdot 10^{-5} T)=3.5 \cdot 10^{-19} N$

6 0

3 years ago

Indique onde em quantos metros o rapaz chegará com as seguintes condições: S0 = 0 a) v = 3 m/s e t = 2 s b) v = 2 m/s e t = 3,5

Naily [24]

Answer:

I will answer in English.

Here we will use the relation

Velocity*time = distance

So:

a) velocity = 3m/s

time = 2s

Distance = 3m/s*2s = 6m

b) velocity = 2m/s

time = 3.5s

Distance = 2m/s*3.5s = 7m

c) velocity = 10m/s

time = 0.5s

Distance = 10m/s*0.5s = 5m

d) velocity = 4m/s

time = 2.5s

Distance = 4m/s*2.5s = 9m

e) velocity = 1.5m/s

time = 5s

Distance = 1.5m/s*5s = 7.5m

8 0

3 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

The ball has charge $-q_0$ , and the ring has positive charge $+Q$ distributed uniformly along its circumference.

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ (we use this approximation instead of the original $d^2+a^2\approx a^2$ since $z$ , our assumption still holds )

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

5 0

4 years ago

A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 13.5 m/s

Yakvenalex [24]

The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is 73.5°.

<h3>What is free falling?</h3>

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

where v is the speed just before hitting the ground.

A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt (v² - 2gh)

u = sq rt (13.5² - 2x9.81x4.4)

u = 9.798 m/s or 9.8 m/s

Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,