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Rina8888 [55]
2 years ago
8

A rotating beacon is located 2 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the bea

con rotates at 10 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving at an instant when the beam is lighting up a point 2 miles along the shore from the point A?
Physics
1 answer:
Mice21 [21]2 years ago
6 0

Answer:

v = 3369.2 m/s

Explanation:

As we know that Beacon is rotating with angular speed

f = 10 rev/min

so we have

\omega = 2\pi f

\omega = 2\pi(\frac{10}{60})

\omega = 1.047 rad/s

now we know that

v = r \omega

here we will have

r = 2 miles = 2(1609 m)

r = 3218 m

so we have

v = 3218(1.047)

v = 3369.2 m/s

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Given:
rod of circular cross section is subjected to uniaxial tension.
Length, L=1500 mm
radius, r = 10 mm
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Force, F=20 kN = 20,000 N
[note: newton (unit) in abbreviation is written in upper case, as in  N ]

From given above, area of cross section = π r^2 = 100 π =314 mm^2
(i) Stress,
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(iii) elongation
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Answer:

C) The pressure reading stays the same.

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A 0.050 kg bullet strikes a 5.0 kg wooden block with a velocity of 909 m/s and embeds itself in the block which fies off its sta
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mv=(m+M)V\\\\V=\dfrac{mv}{m+M}\\\\V=\dfrac{0.05\times 909}{0.050+5}\\\\V=9\ m/s

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Explanation:

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In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.  She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

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t = 15 min = 0.25 hour

v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h

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