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2 years ago

When the forces acting on a particle are resolved into cylindrical components, friction forces always act in the

Physics

1 answer:

Aleksandr [31]2 years ago

7 0

Answer:

Tangential

Explanation: This is a kind of force which act on a moving body in such a way that it is curved in the direction of the path of the body. This implies that when the velocity of the object is positive, the acceleration will be negative.

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In Millikan's oil drop experiment an oil drop is at rest between two large plates separated by 1.3 cm when the potential differe

sweet-ann [11.9K]

Answer:

Mass of the oil drop, $m=3.01\times 10^{-15}\ kg$

Explanation:

Potential difference between the plates, V = 400 V

Separation between plates, d = 1.3 cm = 0.013 m

If the charge carried by the oil drop is that of six electrons, we need to find the mass of the oil drop. It can be calculated by equation electric force and the gravitational force as :

$qE=mg$

$m=\dfrac{qE}{g}$

$q=6e$ , e is the charge on electron

E is the electric field, $E=\dfrac{V}{d}=\dfrac{400}{0.013}=30769.23\ V/m$

$m=\dfrac{6\times 1.6\times 10^{-19}\times 30769.23}{9.8}$

$m=3.01\times 10^{-15}\ kg$

So, the mass of the oil drop is $3.01\times 10^{-15}\ kg$ . Hence, this is the required solution.

5 0

3 years ago

Two wheels with fixed hubs and radii 0.51 m and 1.9 m, each having a mass of 3 kg, start from rest. Forces 5 N and F2 are applie

Katarina [22]

Answer:

18.63 N

Explanation:

Assuming that the sum of torques are equal

Στ = Iα

First wheel

Στ = 5 * 0.51 = 3 * (0.51)² * α

On making α subject of formula, we have

α = 2.55 / 0.7803

α = 3.27

If we make the α of each one equal to each other so that

5 / (3 * 0.51) = F2 / (3 * 1.9)

solve for F2 by making F2 the subject of the formula, we have

F2 = (3 * 1.9 * 5) / (3 * 0.51)

F2 = 28.5 / 1.53

F2 = 18.63 N

Therefore, the force F2 has to 18.63 N in order to impart the same angular acceleration to each wheel.

3 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$