A train moved forward at the speed of 38m / s and whistled 600m away from a tunnel entrance. When the driver heard the echo, how far was the locomotive from the tunnel entrance? (the sound velocity is 340m / s) (do not set the unknown number method to solve, but answer in the way of known, seek and answer.)

A train moved forward at the speed of 38m / s and whistled 600m away from a tunnel entrance. When the driver heard the echo, how far was the locomotive from the tunnel entrance? (the sound velocity is 340m / s) (do not set the unknown number method to solve, but answer in the way of known, seek and answer.)

490m step by step, easy to understand. Suppose the echo is heard after T seconds. At this time, the distance traveled by the sound + the distance traveled by the car = 600 + 600. Note that the actual speed of the sound at this time is sound speed + vehicle speed, that is, 340 + 38 (vehicle speed + actual sound speed) t = 600 + 600 (38 + 340 + 38) t = 1200t = 1200 / 416, and the distance from the tunnel entrance = 600-38 * 1200 / 416 = 490 (m)

A train is moving forward at the speed of 20m / s, and the driver whistles 500m away from the tunnel entrance. When an echo is heard, how far is the locomotive from the tunnel?

Suppose that when the echo is heard, the distance between the locomotive and the tunnel is s, and the distance through which the train passes is 500m-s, from v = s
T, the time taken is T1 = S1
v1=500m−s
20m/s；
Then the distance of sound passing is s + 500m and the time taken is T2 = S2
v2=500m+s
340m / s, according to the meaning of the question, the sound is equal to the time taken by the train, then:
500m−s
20m/s=500m+s
340m/s
Solution: s ≈ 444.4m
A: when the echo is heard, the distance between the locomotive and the tunnel is about 444.4m

A 150m long train passes through a 600m tunnel at the speed of 15m / kg. It takes - hours from the locomotive entering the tunnel entrance to the tail of the train leaving the tunnel

(150+600)/15=

Log and LG in a logarithmic function. Is the latter a logarithm based on ten

Yes, the latter is the abbreviation of logarithm based on ten! The watchtower owner accepts it. If you don't understand it, you can continue to ask

Application of formula A ^ log (a) n = n 3 ^ log (9) (lg2-1) ^ 2 + 5 ^ log (25) (lg0.5-2) ^ 2 is equal to?

3^log(9)(lg2-1)^2 + 5^log(25)(lg0.5-2)^2 =3^[1/2log3 (lg2-1)^2]+5^[1/2log5 (lg0.5-2)^2]=3^log3 (1-lg2)+5^log5 (2-lg0.5)=1-lg2+2-lg0.5=3-(lg2+lg0.5)=3-lg1=3

Let a and B be the two real roots of square LG (x square) - 3 = 0 of equation (lgx), and find that the log bottom is the value of B on a + the log bottom is the value of a on B

(lgx) ²－ 2lgx-3 = 0 (lgx-3) (lgx + 2) = 0, lgx = 3 or lgx = - 2
A = 1000 B = 1 / 100 or a = 1 / 100 B = 1000
When a = 1000 B = 1 / 100, ㏒ a B = lg1000 1 / 100 = lg1000 / LG (1 / 100) = - 3 / 2
㏒a b＋ ㏒b a＝㏒a b＋1/㏒a b＝﹣3/2＋(-2/3) ＝﹣13/6
When a = 1 / 100 B = 1000, ㏒ a B = LG1 / 100 1000 = - 2 / 3
㏒a b＋ ㏒b a＝㏒a b＋1/㏒a b＝﹣2/3＋(-3/2) ＝﹣13/6