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frosja888 [35]
3 years ago
13

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

uring 30 seconds of rotation? ​ ​
Physics
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

\theta = \omega t\\\\\theta = (100 \frac{rev}{\min}  \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi  \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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To gather more information and details on the soil
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3 years ago
Escribo dentro del paréntesis la letra (F) si el enunciado es falso o (V) si el enunciado es verdadero. Fundamento si mi elecció
KatRina [158]

Answer:

El primero es (V) y el segundo es(f)

4 0
3 years ago
A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is
Kruka [31]

Answer:

3.76 m/s

Explanation:

Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.

v' = dx(t)/dt..................... Equation 1

Where v' = instantaneous velocity, x = distance, t = time.

Given the expression,

x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³

x(t) = 28 + 12.4t - 0.0450t³

Differentiating x(t) with respect to t.

dx(t)/dt = 12.4 - 0.135t²

dx(t)/dt = 12.4 - 0.135t²

When t = 8.00 s.

dx(t)/dt = 12.4 - 0.135(8)²

dx(t)/dt = 12.4 - 8.64

dx(t)/dt = 3.76 m/s.

Therefore,

v' = 3.76 m/s.

Hence, the instantaneous velocity = 3.76 m/s

8 0
3 years ago
What waves have high amplitudes
Olegator [25]

-- loud sounds

-- bright lights

-- strong radio signals

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-- a tsunami in the ocean

-- earthquakes above Richter 5 or 6

5 0
3 years ago
When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ
pantera1 [17]

Answer:

The period is T =  0.700 \ s

Explanation:

From the question we are told that  

    The mass is m =  0.350  \ kg

     The extension of the spring is  x =  12.0 \ cm = 0.12 \ m

       

The spring constant for this is mathematically represented as

       k  = \frac{F}{x}

Where F is the force on the spring which is mathematically evaluated as

       F  =  mg  =  0.350 * 9.8

       F  =3.43 \ N

So  

    k  = \frac{3.43 }{ 0.12}

    k  = 28.583 \ N/m

The period of oscillation is mathematically evaluated as

      T =  2 \pi \sqrt{\frac{m}{k} }

substituting values

     T =  2  *  3.142*  \sqrt{\frac{0.35 }{28.583} }

     T =  0.700 \ s

   

7 0
3 years ago
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