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1. ) A bug is observed to travel 46 cm in 2.4 seconds. What is the bug's speed in miles per hour?

2. ) A lobster boat is situated due west of a lighthouse. A barge is 12 km south of the lobster boat. From the barge the bearing to the lighthouse is 63 degrees (12 km is the length of the side adjacent to the 63 degree bearing). How far is the lobster boat from the light house?

3. ) A recent land survey was conducted on a vacant lot where a commercial building is to be erected. The plans for the future building construction call for a building having a roof supported by two sets of beams. The beams in the front are 8 feet high and the back beams are 6.5 feet high. The distance between the front and back beams is 8 feet. At what angle will the roof lay on the front beam?

2006-09-12 03:26:42 · 3 answers · asked by dudekunle 3 in Education & Reference Homework Help

3 answers

1. ) A bug is observed to travel 46 cm in 2.4 seconds. What is the bug's speed in miles per hour?

46 cm * 6.21371192 × 10^-6 cm / 1 mi = 0.0002858 mi
2.4 sec * 1 hr / 3600 sec = 0.0006667 hr

0.0002858 mi / 0.0006667 hr = .43 mi/hr

2. ) A lobster boat is situated due west of a lighthouse. A barge is 12 km south of the lobster boat. From the barge the bearing to the lighthouse is 63 degrees (12 km is the length of the side adjacent to the 63 degree bearing). How far is the lobster boat from the light house?

This is represented by a right triangle. Let a = 63 degrees. tan a = opposite/adjacent.
tan 63 = x/12
x = (tan 63)/12 = 23 km

3. ) A recent land survey was conducted on a vacant lot where a commercial building is to be erected. The plans for the future building construction call for a building having a roof supported by two sets of beams. The beams in the front are 8 feet high and the back beams are 6.5 feet high. The distance between the front and back beams is 8 feet. At what angle will the roof lay on the front beam?

Set it up as a triangle. The adjacent side is 8 feet, and the opposite side is 1.5 feet (the difference in the beams.

tan a = 1.5/8 = 3/16
a = 10.6 degrees

2006-09-12 03:46:04 · answer #1 · answered by ³√carthagebrujah 6 · 0 0

For questions 2 and 3 I would draw diagrams of what is being described. It would really help you to learn how to do these mechanics problems yourself, but perhaps with some quidance. If your university has a tutorial center I would go there after you had thoroughly tried them yourself. You could also ask a friend from class to walk you through it. These are also pretty much geometry problems (except #1 is just conversion). You could try looking in an old geometry text for the concepts. I think you'll find the coming test and final exam to be easier if you do this.

2006-09-12 03:38:16 · answer #2 · answered by FairyMaryGirl 2 · 0 0

I would answer all of these if they were three separate questions. In other words it is not worth a mere ten points; it is worth thirty.

So here is your one answer:

1.) 46cm/2.4sec*3600 sec/hr = 69000 cm/hr = 0.428746123 mph

2006-09-12 03:30:56 · answer #3 · answered by jimvalentinojr 6 · 0 0

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