As shown in the figure, the typhoon center measured by the meteorological station of city a is at B, 600km to the west of city a, and the speed of 200km / h is 60 ° to the north and East The area 500 km away from the center of the typhoon is affected by the typhoon. (1) Is city a affected by the typhoon? Why? (2) If a Cheng is affected by this typhoon, how long will city a be affected by this typhoon?

As shown in the figure, the typhoon center measured by the meteorological station of city a is at B, 600km to the west of city a, and the speed of 200km / h is 60 ° to the north and East The area 500 km away from the center of the typhoon is affected by the typhoon. (1) Is city a affected by the typhoon? Why? (2) If a Cheng is affected by this typhoon, how long will city a be affected by this typhoon?

(1) Through point a, ad is perpendicular to point BC and perpendicular to point D. in right triangle ADB, ∠ abd = 30 ° so ad = AB / 2 = 300,
So city a must be affected by the typhoon
(2) Take a as the center of the circle, 500 as the radius, make the intersection line BC at points E and F, when the typhoon center is between E and F, city a is affected by the typhoon. Connect AE, in the right angle △ ade, AE = 500, ad = 300, then de = 400, thus EF = 800, the moving speed of the typhoon center is 200, so city a is affected by the typhoon for 800 / 200 = 4 hours

As shown in the figure, the typhoon center of city a meteorological station is located at B, 600km to the west of city a, and moves towards BF, 60 ° to the East, at the speed of 200km per hour
500 km from the center of the typhoon will be affected
1. Will city a be affected? Why
2. If it is affected, how long will it be affected? Picture C can be changed to F. to solve it with euroshare theorem, I can't listen to the teacher when I am sick. Although it is an additional question of the test paper, I must master it

1﹚
In RT ⊿ AMB, ∠ ABM = 30 & # 186;, ab = 600km  am = &# 189; ab = 300km ＜ 500km
A city will be affected
2) ∵ when the distance d between typhoon center F and city a is less than 500 km, city a will be affected
(1) when the typhoon center f is on the line am, FA = 500km, it begins to affect city A
In RT ⊿ FMA, FA = 500, Ma = 300 ∪ from Pythagorean theorem, MF = 400km
② When the typhoon center f ′ is on the ray MC, f ′ a = 500km, the typhoon starts to leave city A
In this case, in RT ⊿ f ′ Ma, f ′ a = 500, Ma = 300 ∪ from Pythagorean theorem, MF ′ = 400km
When the center of typhoon moves from F to f '(FF ′ = 800km), city a will be affected for 4 hours (800 / 200)

As shown in the figure, the typhoon center measured by the meteorological station of city a is at B, 600km to the west of city a, and moves towards BF, 60 ° to the north by East, at the speed of 200km per hour（
1) Is city a affected by the typhoon? Why?
(2) If city a is affected by this typhoon, how long will city a be affected by this typhoon?

1, yes! Due to the influence of the northern hemisphere cyclone, it rotates counterclockwise from all sides to the center!
two

The center of a typhoon is located in O. the center of the typhoon moves to the northwest at a speed of 25 km / h, and will be affected within a radius of 240 km. City a is 320 km away from O due west of O. will city a be affected by the typhoon? If affected, how many hours?

As shown in the figure, OA = 320, ∠ AON = 45 °, make a vertical line of on through point a, the perpendicular foot is h, take a as the center of the circle, 240 as the radius, draw the arc intersection line Oh to m, N, in RT △ OAH, ah = oasin45 ° = 1602 < 240, so a city will be affected, in RT △ AHM, MH = AM2 − ah2 = 2402 − (1602) 2 = 80 ∧ Mn = 160, affected

The total length of a certain section of high-speed railway is 1775km, and the preset running speed of "harmony" EMU can reach 355km / h. how many hours will it take to pass through this section of high-speed railway?
Through this section of high-speed rail EMU, the average speed of the express train can be calculated by 9.5h less than that of the fast train?

The total length of a high-speed railway section is 1775km, and the preset running speed of "harmony" EMU can reach 355km / h. how many hours does it take to pass this section of high-speed railway? S = 1775km, v = 355km / h is obtained from v = s / T. the time required for "harmony" EMU to pass this section of high-speed railway is t = s / v = 1775km / 355km / h = 5h

1. The preset running speed of the EMU can reach 200km / h to 250km / h, and the total length of the EMU is 300km. It used to take 4 hours from Haikou to Sanya, but now it only takes 1 hour and 40 minutes

Hello
1 hour 40 minutes = 1 and 2 / 3 hours = 5 / 3 hours
Average speed 300 △ 5 / 3 = 180 km / h
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The whole line of Beijing Guangzhou railway was put into operation on December 26, 2012. The running speed of high-speed train is higher than that of current train set [150km / h]
According to this calculation, how many hours does it take to take a high-speed train from Beijing to Guangzhou [the distance is about 2280km]

2280 △ 150 × (1 + 100%) = 2280 △ 300 = 7.60 (hours)

There is a video of a Wuhan Guangzhou high-speed railway EMU shot with a fixed lens. Xiaotian estimates the running speed of the locomotive by playing the video. It is known that the length of the locomotive is s, and the calculation steps are as follows. Please complete step C and arrange the reasonable order of the calculation steps______ (fill in the letters). A. record the time when the locomotive head arrives at the observation point. & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; B. calculate the time taken for the whole train to pass through the observation point T.C. select one on the screen______ D. use v = s / T to calculate the running speed of locomotive. E. record the time when the locomotive tail reaches the observation point

The first step is to select an observation point as a fixed place, the second step is to record the arrival time of the train head, the third step is to record the arrival time of the train tail, the fourth step is to calculate the passing time of the whole train, and the fifth step is to calculate the train speed according to the formula v = St

The first flight is 200 km / h, and then 250 km / h. The first flight is 390 km more than the second flight, with an average speed of 220 km / h. how many km?

An aircraft first flies a distance at a speed of 200 km / h, and then uses 250 km / h to complete the whole flight. It is known that the first distance is 390 km more than the second distance, and the average speed is 215 km / h
Let the first time be T1, the second time be T2, and the whole process be s
V1T1-V2T2=390 200T1-250T2=390 T1=(25T2+39)/20
S / (T1 + T2) = vmean by substituting the above formula into the left formula, we can see that s = vmean * (T1 + T2) = 215 * (20t2 + 25t2 + 39) / 20
At the same time, s = v1t1 + v2t2
It can get 200t1 + 250t2 = 215t1 + 215t2 by juxtaposing with card type
T1=7/3T2
Therefore, it is not necessary to calculate: T2 = 18S, T1 = 42s
S=215*60=12900m
We can get s from the above three formulas

After a helicopter flies from ground a to ground B at the speed of 250km / h, it turns around and flies back to ground a at the speed of 200km / h. It takes 6.75 hours to find the ground a
It's impossible to use the formula. If you use the equation, there will be extra points

Is it the distance between a and B?
x/250+x/200=6.75-------x=750KM