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8090 [49]
4 years ago
15

A floor polisher has a rotating disk that has a 14-cm radius. The disk rotates at a constant angular velocity of 1.3 rev/s and i

s covered with a soft material that does the polishing. An operator holds the polisher in one place for 36 s, in order to buff an especially scuffed area of the floor. How far does a spot on the outer edge of the disk move during this time

Physics
2 answers:
luda_lava [24]4 years ago
6 0

Answer:

Distance moved = 41.2m

Explanation:

Detailed explanation and calculation is shown in the image below

erik [133]4 years ago
4 0

Answer:

d = 41.17 m

The outer edge of the disk moves 41.17m during this time

Explanation:

Given;

angular velocity w = 1.3 rev/s

Radius of rotating disc r = 14 cm = 0.14m

Time t = 36s

The linear velocity v can be derived using;

v = 2πwr

And the distance covered can be expressed as;

d = vt = 2πwrt

d = 2πwrt .......1

Where; w,t,r are angular velocity,time and radius respectively.

Substituting the given values into equation 1

d = 2π×1.3×0.14×36

d = 41.17 m

The outer edge of the disk moves 41.17m during this time

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We want to find the work done and power exerted, let’s start with work first.

We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.

Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2

(The equation says x but can be used with y values)

If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.

To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.

Force = 300 kg * 1.25 m/s^2 = 375 Newtons

So, work = 500 N * 2.5 m = 1000 Joules

Power: W/t

So, Power = 1000 J / 2 seconds = 500 Watts

Hope this helps!

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A flute plays a note with a frequency of 266 Hz. What is the speed of sound if the wavelength is 1.3 m?
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3 years ago
A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 40 m/s at an angle of 37° w
sleet_krkn [62]

Answer:

a) Maximum height = 36.6 m

b) Horizontal distance at which the ball lands = 166.1 m

c) x-component = 32 m/s. y-component = - 27 m/s  

Explanation:

Please, see the attached figure for a description of the problem.

The velocity vector "v" of the cannonball has two components, a horizontal component, "vx", and a vertical component "vy". Notice that at the maximum height, the vertical component "vy" of the velocity vector is 0.

In the same way, the position vector "r" is composed by "rx", its horizontal component, and "ry", the vertical component.

The velocity vector "v" ad the position vector "r" at time "t" are given by the following equations:

v = (v0 * cos α, v0 * sin α + g * t)

r = (x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

Where

v0 = magnitude of the initial velocity vector

α = launching angle

g = gravity acceleration (-9.8 m/s², because the y-axis points up)

t = time

x0 = initial horizontal position

y0 = initial vertical position

If we consider the origin of the system of reference as the point at which the cannonball leaves tha catapult, then, x0 and y0 = 0

a) We know that at maximum height, the vertical component of the vector "v" is 0, because the ball does not move up nor down at that moment (see figure). Then:

0 = v0 * sin α + g * t

-v0 * sin α / g = t

-40 m/s * sin 37° / -9.8 m/s² = t

t = 2.5 s

We can now calculate the position of the cannonball at time t=2.5 s to obtain the maximum height:

r = (x0 + v0 * t cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

The max height is the magnitude of the vector ry max (see figure). The vector ry max is:

ry = (0, y0 + v0 t sin α + 1/2 g * t²)

magnitude of ry = |ry|= \sqrt{(0m)^{2} + (y0 + v0* t*sin \alpha+ 1/2*g*t^{2})^{2}}= y0 + v0*t*sin \alpha + 1/2*g*t^{2})

Then:

max height = y0 + v0 * t * sin α + 1/2 * g * t²

max height = 0 m + 40 m/s * 2.5 s * sin 37° - 1/2* 9.8 m/s² * (2.5 s)² = 29.6 m

Since the ball leaves the catapult 7 m above the ground, the max height above the ground will be 29.6 m + 7 m = 36.6m

<u>max height = 36.6 m</u>

b) When the ball hits the ground, the position is given by the vector "r final" (see figure). The magnitude of "rx", the horizontal component of "r final", is the horizontal distance between the catapult and the wall.

r final = ( x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

We know that the vertical component of "r final" is -7 (see figure).

Then, we can obtain the time when the the ball hits the ground:

y0 + v0 * t * sin α + 1/2 * g * t² = -7 m

0 m + 40 m/s * t * sin 37° + 1/2 g * t² = -7 m

7 m + 40 m/s * t * sin 37° + 1/2 (-9.8 m/s²) * t² = 0

7 m + 24.1 m/s * t - 4.9 m/s² * t² = 0

solving the quadratic equation:

t = 5.2 s (The negative solution is discarded).

With this time, we can calculate the value of the horizontal component of "r final"

Distance to the wall = |rx| = x0 + v0 t cos α

|rx| = 0m + 40 m/s * 5.2 s * cos 37° =<u> 166.1 m</u>

c) With the final time obtained in b) we can calculate the velocity of the ball:

v = (v0 * cos α, v0 * sin α + g * t)

v =(40 m/s * cos 37°, 40 m/s * sin 37°  -9.8 m/s² * 5.2 s)

v =(32 m/s, -27 m)

x-component = 32 m/s

y-component = - 27 m/s

7 0
4 years ago
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