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just olya [345]
2 years ago
13

A common carnival ride, called a gravitron, is a large r = 11 m cylinder in which people stand against the outer wall. the cylin

der spins so that the riders move with a linear speed v. At a certain point the floor of the cylinder lowers and the people are surprised they don't slide down. The friction between the wall and their clothes is μs = 0.62. What is the minimum speed, in meters per second, that the cylinder must make a person move at to ensure they will "stick" to the wall?
Physics
1 answer:
Leya [2.2K]2 years ago
7 0

Answer:

 v = 13.19 m / s

Explanation:

This problem must be solved using Newton's second law, we create a reference system where the x-axis is perpendicular to the cylinder and the Y-axis is vertical

 

X axis

       N = m a

Centripetal acceleration is

       a = v² / r

Y Axis

      fr -W = 0

      fr = W

The force of friction is

     fr = μ N

Let's calculate

    μ (m v² / r) = mg

   μ v² / r = g

   v² = g r / μ

   v = √ (g r /μ)

   v = √ (9.8 11 / 0.62)

   v = 13.19 m / s

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Fynjy0 [20]
The formula for potential energy is PE=mgh
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7 0
2 years ago
What is the initial velocity of the object represented by the graph? ___m/s Graph:
Alex17521 [72]

Answer:

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4 0
2 years ago
Read 2 more answers
before colliding, the momentum of block A is +15.0 kg m/s. after, block A has a momentum -12.0 kg*m/s. what is the momentum of b
Helen [10]

Answer:

The momentum of block B = 27 Kg m/s

Explanation:

Given,

The initial momentum of block A, MU = 15 Kg m/s

The final momentum of block A, MV = -12 Kg m/s

Consider the block B is initially at rest.

Therefore, the initial momentum of block B, mu = 0

According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.

                               <em> MU + mu = MV + mv</em>

                                15  +  (0) = (-12) + mv

                                         mv = 15 + 12

                                              =  27 Kg m/s

Hence, the momentum of the block B after impact is, mv = 27 Kg m/s

3 0
2 years ago
If the loss of 3500 kcal is equal to a loss of 1.0 lb, how many days will it take charles to lose 5.0 lb
rusak2 [61]
We are missing an important piece of information needed to answer this question: the number of kcal Charles losses per day. However, we can come up with a general equation in which kcal/day is the only independent variable.

We know that it takes 3500 kcal to lose one pound. To lose 5 pounds, Charles needs to lose 5 x 3500 kcal = 17,500 kcal. 

To find how many days it takes Charles to lose 17,500 kcal (5 pounds), we must divide that amount by the number of kcal Charles loses per day. 
Here is the equation to calculate that number

Number of days= 17500 / (kcal per day)

If given calories, remember that 1000 calories = 1 kcal, and .001 kcal = 1 cal

4 0
2 years ago
The Palo Verde nuclear power generator of Arizona has three reactors that have a combined generating capacity of 3.937×109 W . H
den301095 [7]

Answer:

t = 2.68 x 10¹⁴ years

Explanation:

First we need to find the amount of energy that Sun produce in one day.

Energy = Power * Time

Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)

Energy of Sun in 1 day = 3.32 x 10³¹ J

Now, the time required by the nuclear power generator, in years, will be:

Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J

3.32 x 10³¹ J = Power * Time

3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)

t = 3.32 x 10³¹ /1.24 x 10¹⁷

<u>t = 2.68 x 10¹⁴ years</u>

8 0
2 years ago
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