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Strike441 [17]
3 years ago
14

If the room radius is 4.5 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out, what is the mini

mum coefficient of static friction so that the people will not slip down?
Physics
1 answer:
kondaur [170]3 years ago
3 0
<span>The force of static friction F equals the coefficient of friction u times the normal force N the object exerts on the surface: F = uN. N is the centripetal force of the wall on the people; N = ma_N, where m is the mass of the people and a_N is the centripetal acceleration. The people will not slip down if F is greater than the force of gravitation: F = uma_N > mg, or u > g/a_N. a_N is the velocity v of the people squared divided by the radius of the room r: a_N = v^2/r. The circumference of the room is 2 pi r = 28.3 m. So v = 28.3 * 0.8 m/sec = 22.6 m/sec. So a_N = 114 m/sec^2. g = 9.81 m/sec^2, so u must be at least 9.81/114 = 0.086.</span>
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A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as t
salantis [7]

Answer:

2.07

Explanation:

Since you didn't supply the drawing, here is what I assumed:

A is the corner opposite the axis of rotation

B is one of the remaining two corners

L1 is the side between A & B

Centripetal acceleration is given by:

ac = v^2 / r = (v / r) * (v / r) * r…………1

Also angular speed is

w = v / r,………….2

Substituting (2) in (1) gives:

ac = (v / r) * (v / r) * r……….3

= (v / r)^2 * r

= w^2 * r

Therefore, the angular acceleration at A and at B are given by:

acA = w^2 * rA……..4

acB = w^2 * rB……..5

It is given that:

acA = n * acB…………6

Substituting (4) and (5) into (6) gives:

w^2 * rA = n * w^2 * rB ……….7==>

rA = n * rB……..8

In terms of the sides L1 and L2:

rA = sqrt (L1^2 + L2^2)…….9

and

rB = L2…………10

Considering (8):

n * L2 = sqrt (L1^2 + L2^2)………11

Squaring both sides:

n^2 * L2^2 = L1^2 + L2^2……….12

Dividing by L2^2:

n^2 = L1^2 / L2^2 + L2^2 / L2^2…….13

= (L1 / L2)^2 + 1 ==>

n^2 - 1 = (L1 / L2)^2 ………14==>

L1 / L2 = sqrt (n^2 - 1) ………15

= sqrt (2.30^2 - 1)

= 2.07. . . . . . <<<=== the value of the ratio L1 / L2 when n = 2.30

8 0
3 years ago
What pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c?
Ronch [10]

The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

Therefore, option A is correct option.

Given,

Mass m = 14g

Volume= 3.5L

Temperature T= 75+273 = 348 K

Molar mass of CO = 28g/mol

Universal gas constant R= 0.082057L

Number of moles in 14 g of CO is

n= mass/ molar mass

= 14/28

= 0.5 mol

As we know that

PV= nRT

P × 3.5 = 0.5 × 0.082057 × 348

P × 3.5 = 14.277

P = 14.277/3.5

P = 4.0794 atm

P = 4.1 atm.

Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.

learn more about pressure:

brainly.com/question/22613963

#SPJ4

5 0
1 year ago
two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel .calculate the equivalent series resist
Novosadov [1.4K]

Explanation:

Given that,

Two resistors of resistance 6 ohm and 3 ohm are connected in series and then in parallel.

For series combination,

R_{eq}=R_1+R_2

For parallel combination,

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}

When 6 ohm and 3 ohm are in series,

R_s=6+3\\\\R_s=9\ \Omega

When 6 ohm and 3 ohm are in paralle,

\dfrac{1}{R_p}=\dfrac{1}{6}+\dfrac{1}{3}\\\\R_p=2\ \Omega

So, the equivalent resistance in series combination is 9 ohms and in parallel combination it is 2 ohms.

6 0
3 years ago
A 12-kg dog jumps up in the air to catch a ball. The dog's center of mass is normally 0.20 m above the ground, and he is 0.50 m
ad-work [718]

Explanation:

Given Data:

mass of dog = 12 Kg

dog's center of mass = 0.20m

length of dog = 0.50m

height of dog's jump = ?

Solution:

Work done of gravitational force = Gain in Potential energy

2.1 × mgΔh = mg (h - 0.1)

2.1 × (0.3 - 0.1) = (h - 0.1)

h = 0.52 m

5 0
3 years ago
In freefall, heavier objects fall with a greater acceleration than lighter objects.
Rom4ik [11]

Answer:

That is not true all objects fall at the same speed excepts things like feathers or paper.

6 0
3 years ago
Read 2 more answers
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