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2 years ago

14

In a "Rotor-ride" at a carnival, people rotate in a vertical cylindrically walled "room." If the room radius is 5.5 m, and the r

otation frequency 0.5 revolutions per second (rps) when the floor drops out, what minimum coefficient of friction keeps the people from slipping down? What type of friction is this? People on this ride said they were "pressed against the wall," however what is really happening? 0.18

1 answer:

Crank2 years ago

6 0

Answer:

0.181

Explanation:

We can convert the 0.5 rps into standard angular velocity unit rad/s knowing that each revolution is 2π:

ω = 0.5 rps = 0.5*2π = 3.14 rad/s

From here we can calculate the centripetal acceleration

$a_c = \omega^2r = 3.14^2*5.5 = 54.3 m/s^2$

Using Newton 2nd law we can calculate the centripetal force that pressing on the rider, as well as the reactive normal force:

$F = N = a_cm = 54.3 m$

Also the friction force and friction acceleration

$F_f = N\mu = 54.3 m \mu N$

$a_f = F_f / m = 54.3 \mu$

For the rider to not slide down, friction acceleration must win over gravitational acceleration g = 9.81 m/s2:

$g = a_f = 54.3 \mu$

$9.81 = 54.3 \mu$

$\mu = 9.81 / 54.3 = 0.181$

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Types of RM radiation that make up the electromagnetic spectrum are microwaves, infrared light, ultraviolet light, X-rays, and g

HACTEHA [7]

Answer:

True

Explanation:

The electromagnetic spectrum can be regarded as term that give description of range of light that are in existence, and this range from radio waves up to gamma rays. It can be explained as

range of frequencies that electromagnetic radiation takes with thier respective wavelengths as well as photon energies

The Types of electromagnetic radiation that make up the electromagnetic spectrum are

✓ microwaves

✓ infrared light

✓ ultraviolet light

✓X-rays,

✓gamma rays.

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3 years ago

A man ties one end of a strong rope 8.17 m long to the bumper of his truck, 0.524 m from the ground, and the other end to a vert

Kamila [148]

Answer:

2442.5 Nm

Explanation:

Tension, T = 8.57 x 10^2 N

length of rope, l = 8.17 m

y = 0.524 m

h = 2.99 m

According to diagram

Sin θ = (2.99 - 0.524) / 8.17

Sin θ = 0.3018

θ = 17.6°

So, torque about the base of the tree is

Torque = T x Cos θ x 2.99

Torque = 8.57 x 100 x Cos 17.6° x 2.99

Torque = 2442.5 Nm

thus, the torque is 2442.5 Nm.

8 0

3 years ago

A playground slide is in the form of an arc of a circle with a maximum height of 3.0 m, with a radius of 8.5 m, and with the gro

julia-pushkina [17]

Answer:

a) $s \approx 6.676\,m$

Explanation:

a) Let assume that the ground is not inclined, since the bottom of the playground slide is tangent to ground. Then, the length of given by the definition of a circular arc:

$s = \frac{\pi}{4}\cdot R$

$s=\frac{\pi}{4}\cdot (8.5\,m)$

$s \approx 6.676\,m$

The bottom of the slide has a height of zero. The physical phenomenon around Dr. Ritchey's daughter is modelled after Principle of Energy Conservation. The child begins at rest:

$U_{g,A} = K_{B} + W_{fr}$

$m\cdot g \cdot h_{A} = \frac{1}{2}\cdot m \cdot v_{B}^{2} + f\cdot s$

The average frictional force is cleared within the expression:

$f = \frac{m\cdot (g\cdot h_{A}-\frac{1}{2}\cdot v_{B}^{2})}{s}$

$f = \frac{(12\,kg)\cdot [(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)-\frac{1}{2}\cdot (4.5\,\frac{m}{s} )^{2} ]}{6.676\,m}$

$f = 34.684\,N$

4 0

2 years ago

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and

slava [35]

Answer:

f=-1380N

Explanation:

A karate master wants to break a board by hitting the board swiftly with his hand. The master's hand has a mass of 0.30 kg, and as it strikes the board, his hand has a velocity of 23.3 m/s. The master contacts the board for 0.0050 seconds

.the concluding part to the question should be

What is the impact force (impulse) on the board?

solution

from the Newton's second law of motion which states that

the rate of change in momentum is directly proportional to the force applied

f=m(v-u)/t

f=0.3(0-23.3)/0.005

f=-1380N

f=force impact

m=mass of the karates master's hand

t=time for the impact

v=0m/s final velocity

u=initial velocity

6 0

3 years ago

In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

$V_{f}=a_{ave} \Delta t$

$V_{f}=(1.80 m/s^{2})(85.6 s)$

Finally:

$V_{f}=154.08 m/s$

8 0

2 years ago