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Temka [501]
4 years ago
7

To throw the discus, the thrower holds it with a fully outstretched arm. Starting from rest, he begins to turn with a constant a

ngular acceleration, releasing the discus after making one complete revolution. The diameter of the circle in which the discus moves is about 1.6 mm.
If the thrower takes 1.0 s to complete one revolution, starting from rest, what will be the speed of the discus at release?
Physics
1 answer:
yawa3891 [41]4 years ago
6 0

Answer:

Explanation:

Time taken to complete one revolution is called time period.

So, Time period, T = 1 s

Diameter = 1.6 mm

radius, r = 0.8 mm

Let the angular speed is ω.

The relation between angular velocity and the time period is

\omega =\frac{2\pi}{T}

ω = 2 x 3.14 = 6.28 rad/s

The relation between the linear velocity and the angular velocity is

v = r x ω

v = 0.8 x 10^-3 x 6.28

v = 0.005 m/s

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What is a point value for an arrow that is on two different colors?
Whitepunk [10]
It depends on what it is closest to but I would say for instance black is 5 points and red is 6 if u land on the line 5.5
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3 years ago
A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba
Marina CMI [18]

Answer:

4.5m/s

Explanation:

Linear speed (v) = 42.5m/s

Distance(x) = 16.5m

θ= 49.0 rad

radius (r) = 3.67 cm

= 0.0367m

The time taken to travel = t

Recall that speed = distance / time

Time = distance / speed

t = x/v

t = 16.5/42.5

t = 0.4 secs

tangential velocity is proportional to the radius and angular velocity ω

Vt = rω

Angular velocity (ω) = θ/t

ω = 49/0.4

ω = 122.5 rad/s

Vt = rω

Vt = 0.0367 * 122.5

Vt =4.5 m/s

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3 years ago
A sleigh weighing 2000 newtons is pulled my a horse a distance of 1.0 kilometer (or 1000 meters) in 45 minutes. what is the powe
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3 years ago
hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body
bagirrra123 [75]

Answer:

(a)

\lambda _{m}=9.332 \times 10^{-6}m

(b)

\lambda _{m}=1.632 \times 10^{-6}m

(c) \lambda _{m}=4.988 \times 10^{-7}m

 

Explanation:

According to the Wein's displacement law

\lambda _{m}\times T = b

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{310}

\lambda _{m}=9.332 \times 10^{-6}m

(b) T = 1500°C = 1500 + 273 = 1773 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}

\lambda _{m}=1.632 \times 10^{-6}m

(c) T = 5800 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}

\lambda _{m}=4.988 \times 10^{-7}m

5 0
3 years ago
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