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A Street Light Is Mounted at the Top of a 15-ft-tall Pole. a Man 6 Ft Tall Walks Away ...?

we will have 2 triangles: the big triangle, top of light pole "A" , base "B" and the tip of the shadow "T" and the small one, the height of the runner with head at A', feet at B' and T the tip of the shadow the triangles are proportional, 6/15 = (AT - 45) / A T AT = 75 ft S = v* t the guy spared t = S/v = 45/7 = 6.43 seconds same time spend by the tip of the shadow, but the space (") is bigger 75 = v * t = v*6.43 ===> v = 75/6.43 = 11.66 ft/s 1. WHY DOES THIS STREET LIGHT TURN OFF/ON?

Some loose connection

2. i Just Shot a Street light would anything happen to me?

probably charged with misdemeanor destruction of property, charged hefty fine and released to the custody of your parents 3. I have a street light that needs a power source. I'd like to use solar instead of hard wire. What system should I use?

Q: I have a street light that needs a power source. I would like to use solar instead of hard wire. What system should I use? A: One that can accumulate and store in a battery on the shortest darkest day of the year enough energy to illuminate the entire night which is going to be the longest night of the year.

4. A woman of height of 6 ft walks at a rate of 7ft/s towards a point P directly beneath a street light that is 30 ft above the ground .What is the rate of change of the distance between the tip of her shadow and the point P?

Let the length of the woman's shadow = xand let her distance from the street light = yThen by similar triangles, we have6/x = 30/(x y)6(x y) = 30xx y = 5xy = 4xThe tip of the shadow is always (x y)/y = 5/4 times as far as the woman from the point P, so it must be moving at 5/4 times her speed.Thus the rate of change of distance of the tip of her shadow from the point P (or its speed) is7 ft/s * (5/4) = 8 3/4 ft/s.A woman of height of 6 ft walks at a rate of 7ft/s towards a point P directly beneath a street light that is 30 ft above the ground . What is the rate of change of the distance between the tip of her shadow and the point P?

5. A 6 foot man walks at 8 ft per second toward a street light that is 20 ft abovethe ground (cont.)?

nicely in case you draw this out, you will desire to be conscious that they are comparable triangles. One is 6ft (top) and 20ft(length 50-30=20) it extremely is the triangle from the sunshine source to the guy. permit x continually be the present top of the shadow. the 2d triangle is the somewhat construction to the sunshine source. This triangle is x ft severe it extremely is somewhat the top of the shadow, and 50ft long. The triangles are comparable by way of having 3 comparable angles(I talked approximately because it AAA similarity) remedy for x 6/20 = x/50 20x = 3 hundred x = 15ft So at first, the shadow is 15ft tall on the construction. next, I permit the guy flow 5ft in direction of the construction. So your first triangle is now 6ft for top of and 25ft for length the 2d triangle is x for top of the shadow and 25ft for length remedy for x returned 6/25 = x/50 25x = 3 hundred x = 12ft So the shadow is now 12ft tall. finding at the two heights, the shadow is shrinking at a value of 3m/s I even have faith that this shrinking is linear. verify to work out if its linear bypass yet another 5ft First triangle 6ft and 30ft 2d triangle x and 50ft 6/30 = x/50 5x = 50 x = 10ft Dam lol so the equation is not linear. So i am getting a table of values now. 6/35 = x/50 35x = 3 hundred x = approximately 8. 6 it extremely is it for me immediately. i am going to calculate the deceleration of the shadow after a good nights sleep

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