Nama khas matahari adahah?

Lena [83]

When soccer players run they are using friction to propell themselves

7 0

3 years ago

what equastion do you use to solve Riders in a carnival ride stand with their backs against the wall of a circular room of diame

Hitman42 [59]

Answer:

μsmín = 0.1

Explanation:

There are three external forces acting on the riders, two in the vertical direction that oppose each other, the force due to gravity (which we call weight) and the friction force.
This friction force has a maximum value, that can be written as follows:

$F_{frmax} = \mu_{s} *F_{n} (1)$

where μs is the coefficient of static friction, and Fn is the normal force,

perpendicular to the wall and aiming to the center of rotation.

This force is the only force acting in the horizontal direction, but, at the same time, is the force that keeps the riders rotating, which is the centripetal force.
This force has the following general expression:

$F_{c} = m* \omega^{2} * r (2)$

where ω is the angular velocity of the riders, and r the distance to the

center of rotation (the radius of the circle), and m the mass of the

riders.

Since Fc is actually Fn, we can replace the right side of (2) in (1), as

follows:

$F_{frmax} = m* \mu_{s} * \omega^{2} * r (3)$

When the riders are on the verge of sliding down, this force must be equal to the weight Fg, so we can write the following equation:

$m* g = m* \mu_{smin} * \omega^{2} * r (4)$

(The coefficient of static friction is the minimum possible, due to any value less than it would cause the riders to slide down)

Cancelling the masses on both sides of (4), we get:

$g = \mu_{smin} * \omega^{2} * r (5)$

Prior to solve (5) we need to convert ω from rev/min to rad/sec, as follows:

$60 rev/min * \frac{2*\pi rad}{1 rev} *\frac{1min}{60 sec} =6.28 rad/sec (6)$

Replacing by the givens in (5), we can solve for μsmín, as follows:

$\mu_{smin} = \frac{g}{\omega^{2} *r} = \frac{9.8m/s2}{(6.28rad/sec)^{2} *2.5 m} =0.1 (7)$

5 0

2 years ago

An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n)

maria [59]

Answer:

ion

Explanation:

4 0

3 years ago

Read 2 more answers

4. Assume a multiple level queue system with a variable time quantum per queue, and that the incoming job needs 50ms to run to c

Troyanec [42]

Answer:

Explanation:

For the completion of incoming job it will take 50ms

First queue takes 5ms quantum time and the subsequent queue takes double of the previous question

So,

First queue T_1 = 5ms

Second queue T_2 = 2 × T_1 = 2 × 5 = 10ms

Third queue T_3 = 2 × T_2 = 2 × 10 = 20ms

Fourth queue T_4 = 2 × T_3 = 2 × 20 = 40ms

Fifth queue T_5= 2 × T_4 = 2 × 40 = 80ms.

Now, the job will be done after the fifth queue.

So, after the first queue, the job is not completed, so, we have first interruption

After the second queue, the job is not completed, so, we have second interruption

After the third queue, the job is not completed, so we have third interruption.

After the fourth queue, the job is not yet completed, so we have the fourth interruption

And in the fifth queue the job is completed, so we don't have any interruption here.

So, the job will be interrupted 4 times and it will finished on the fifth queue.

6 0

3 years ago

Forces normal to a particle's displacement do no work.

snow_tiger [21]

Answer:

Explanation:

The work done is defined as the product of force applied in the direction of displacement and the displacement.

W = F x d x Cosθ

where, F is the force applied, d be the displacement and θ be the angle between the displacement and force.

For the normal forces, the angle between the displacement and the force applied is 90 degree, and the value of Cos 90 is zero, so the work done is zero.

3 0

3 years ago

Nama khas matahari adahah?​

Nama khas matahari adahah?