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just olya [345]
3 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]3 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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A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?
Serhud [2]

Answer:

<em>The y component of his displacement is 11.22 meters</em>

Explanation:

<u>Components of the displacement</u>

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

x=r\cos\theta

The vertical component is calculated by:

y=r\sin\theta

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:

y=20.51\sin(33.16^\circ)

y=11.22\ m

The y component of his displacement is 11.22 meters

7 0
3 years ago
How can astrophysicists tell whether a star is receding from or approaching earth?
MrMuchimi

Answer:

Doppler shift of the starlight

Explanation:

To predict the movement of a star, we compare the spectra of elements found in star (H, He Na etc.), first spectra which are obtained from star and second spectra from laboratory. If spectral lines of the spectra obtained from star, are shifting towards red end (called red shift) then star is going away from earth and if shifting is towards blue (called blue shift), then star is approaching the earth. This is Doppler's shift.

3 0
3 years ago
a force is applied to a box of 10.0 kg for 4.0 s. the box goes from rest to 25 m/s in that time. What is the magnitude of that f
Paul [167]

Given:

m(mass of the box)=10 Kg

t(time of impact)=4 sec

u(initial velocity)=0.(as the body is initially at rest).

v(final velocity)=25m/s

Now we know that

v=u+at

Where v is the final velocity

u is the initial velocity

a is the acceleration acting on the body

t is the time of impact

Substituting these values we get

25=0+a x 4

4a=25

a=6.25m/s^2

Now we also know that

F=mxa

F=10 x6.25

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8 0
3 years ago
Which equation describes the line containing the points (-2, 3) and (1, 2)​
snow_lady [41]

Answer:

y =  \frac{ - 1}{3}x +  \frac{7}{3}

Explanation:

\frac{y - 3}{2 - 3}  =  \frac{x + 2}{1 + 2}  \\  \ - y + 3 =  \frac{x + 2}{3}  \\  y =   \frac{ - 1}{3}x  +  \frac{7}{3}

8 0
3 years ago
When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le
77julia77 [94]

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

x+2.93

Both cover a magnitude of 8.32 ft, therefore

x +(x+2.97) = 8.32

Now solving for x we have,

x + (x + 2.93) = 8.32

2x + 2.93 = 8.32

2x = 8.32 - 2.93

x = \frac{ 8.32 - 2.93}{2}

x = 2.695 ft

Therefore the shorter stick is 2.695ft long.

7 0
4 years ago
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