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3 years ago

Write how you would use numbers to investigate each object

Physics

1 answer:

Nata [24]3 years ago

5 0

For the 1st imagine you can use numbers such as to count the eyes or the legs. The second image you could count how many circles are in the image. The third one you could use numbers to count the wheels.

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A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part

Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

Polar moment of Inertia

$(I_p)s = \frac{\pi(0.55)4}2$

= 0.14374 in 4

Maximum sustainable torque on the solid circular shaft

$T_{max} = T_{allow} \frac{I_p}{r}$

= $(14 \times 10^3) \times (\frac{0.14374}{0.55})$

= 3658.836 lb.in

= $\frac{3658.836}{12}$ lb.ft

= 304.9 lb.ft

Maximum sustainable torque on the tubular shaft

$T_{max} = T_{allow}( \frac{Ip}{r})$

= $(14 \times10^3) \times ( \frac{0.13101}{0.55})$

= 3334.8 lb.in

= $(\frac{3334.8}{12} )$ lb.ft

= 277.9 lb.ft

Maximum sustainable power in the solid circular shaft

$P_{max} = 2 \pi f_T$

= $2\pi(2.1) \times 304.9$

= 4023.061 lb. ft/s

= $(\frac{4023.061}{550})$ hp

= 7.315 hp

Maximum sustainable power in the tubular shaft

$P _{max,t} = 2\pi f_T$

= $2\pi(2.1) \times 277.9$

= 3666.804 lb.ft /s

= $(\frac{3666.804}{550})$ hp

= 6.667 hp

7 0

3 years ago

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

4 years ago

The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func

Kitty [74]

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = $570 \times \dfrac{2\pi}{60}$

= 59.69 rad/s

1600 rpm = = $570 \times \dfrac{2\pi}{60}$

= 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt

167.6 = 59.7 + α x 7980

α x 7980 = 107.9

α = 0.0135 rad/s²

8 0

3 years ago

A certain radar installation tracks airplanes by transmitting electromagnetic radiation of wavelength 4.0 cm.

Anna35 [415]

the wavelength equation is

speed (of light in this case)= wavelength (m) x frequency

3x10^8m/s / .07m = f

frequency= 4 285 714 286 hertz

b) Total distance= 4.8 km (4,800 m)

Speed = 3x10^8 m/s

d=st

t= d/s

t= 4,800 m/3x10^8m/s

t= 1x10^-5 seconds

3 0

3 years ago

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The distance of an illuminated object from its light source is decreased to 1/3 while the luminous flux remains constant. Which

Nostrana [21]

Answer:

The illuninance being multiplied by 9

Explanation:

6 0

3 years ago

Read 2 more answers