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The ratio of the length of the hour hand, minute hand and second hand on an accurate clock is 2:3:3. Try to find the ratio of the angular velocity of the three hands and the linear velocity of the tip of the three hands Speed up. Thank you······
The clockwise angular velocity is 360 / (12 * 60 * 60) = 1 / 120
The minute hand angular velocity is 360 / (60 * 60) = 1 / 10
The second hand angular velocity is 360 / 60 = 60
The ratio of the three is 1 / 120:1 / 10:60
Length times angular velocity equals linear velocity
Let's figure it out for ourselves
The ratio of the angular velocity of the second hand, minute hand and hour hand is
The second hand cycle is 60s, the minute hand is 3600s, and the hour hand is 3600 times 60s
Therefore, the ratio of their angular velocity is 3600:60:1, which is the inverse ratio of the period
If the length of the minute hand on the clock is 1.5 times that of the hour hand, how many times is the angular velocity of the minute hand and how much is the linear velocity of the end of the minute hand
If the minute hand moves 360 degrees an hour, then the minute is 360 / 60 = 6 degrees, and the hour hand is 360 / 12 = 30 degrees, then the minute is 30 / 60 = 0.5 degrees
Therefore, the angular velocity of the minute hand is 6 / 0.5 times that of the hour hand
Then, according to v = RW, the linear velocity of minute hand is 12 * 1.5 = 18 times that of hour hand
For an accurate clock, the minute hand and the hour hand have a length ratio of 1.2:1,
Why is angular velocity certain?
If the angular velocity is determined, the minute hand turns 360 an hour, and the hour hand turns 30 an hour, that is 12:1. The linear velocity is the angular velocity multiplied by the length, so the ratio is 72:5 or 14.4:1
The hour hand and minute hand of a mechanical watch can be regarded as rotating at a constant speed. How long does it take for the hour hand and minute hand to coincide from the first time to the second time?
According to ω = 2 π T, the ratio of the angular velocity of the hour hand to the minute hand is 1:12. When the hour hand and the minute hand overlap from the first time to the second time, t + 2 π = ω min T. then t = 2 π ω min − ω H = 2 π 1112 ω min = 2 π 11122 π t min = 1211 H The time of treatment was 1211 H
When does the minute hand catch up with the hour hand for the first time? When is the minute hand in line with the hour hand for the first time?
A mathematical problem
The minute hand turns 6 ° in 1 minute, and the hour hand turns 0.5 ° in 1 minute
Let the minute hand catch up with the hour hand for the first time in X minutes
6x-0.5x=360
5.5x=360
X = 720 / 11 minutes ≈ 65.5 minutes = 6 hours, 5 minutes and 30 seconds
Let the minute hand be in line with the hour hand for the first time after y minutes
6x-0.5x=180
5.5x=180
X = 360 / 11 minutes ≈ 32.7 minutes = 32 minutes 42 seconds
The minute hand and second hand of the mechanical watch (as shown in the figure) overlap from the second time to the second time, and the time between them is ()
A. 5960minB. 1 minC. 6059minD. 6160min
The period of minute hand is 1H, the period of second hand is 1min, and the cycle ratio of the two is T1: T2 = 60:1. From the first time to the second time, the coincidence of minute hand and second hand is: ω 1t + 2 π = ω 2T, that is, 2 π t1t + 2 π = 2 π T2T, and T1 = 60t2 = 60min, so t = 6059min, so select: C
The hour hand and the minute of a clock point between 4 and 6, and the "5" on the clock face is just in the middle of the hour hand and the minute?
(1) If the hour hand is between 4 and 5, and the minute hand is between 5 and 6, let this time be x, 12 o'clock be 0 degree, the degree indicated by the hour hand + the degree indicated by the minute hand = 2 times of the degree of 5 o'clock, the equation is & nbsp; 30x + 360 × (x-4) = 2 × 150, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 390x = 1740, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; &If the minute hand is between 4 and 5, and the hour hand is between 5 and 6, then 30x + 360 × (X-5) = 2 × 150390, x = 2100, x = 7013, x = 5513, that is, 5:23:5
Suppose that the mechanical watch moves evenly. On the dial, when one point passes, the hour hand and the minute hand will be combined into a straight line again. As a result, the fraction is retained. How is the calculation method?
It turns into a pursuit problem. Starting from one o'clock, the hour hand is at 1, the minute hand is at 12, and the minute hand starts to pursue. The speed of the hour hand is one grid per hour, and the speed of the minute hand is 12 grids per hour. They are one grid apart. Therefore, the pursuit time is 1 (12-1) = 0.0909 An hour is equivalent to 5.4545 Minutes, coincidence
As shown in the figure, the horizontal metal disk is placed in a uniform magnetic field with magnetic induction intensity of B and vertical downward direction. As the metal rotating shaft DD 'rotates clockwise with angular velocity ω, the center and edge of the disk are respectively connected with the primary coil of an ideal transformer by metal slides The coil is connected to a load with resistance R. if the resistance of the disc and the conductor is neglected, the following statement is correct ()
A. The voltage at both ends of the transformer primary coil is B ω r2b. The voltage at both ends of the transformer primary coil is 12b ω r2c. The current through load R is B ω r22nr D. the current through load R is B ω r2nr
A. The induced electromotive force generated by cutting the magnetic induction line with the radius of the disk: e = 12br2 ω, the voltage at both ends of the original coil of the transformer is 12br2 ω, so a is wrong and B is correct; C. The electromotive force generated by the rotation process of the disk is a fixed value, the magnetic flux through the secondary coil of the transformer is unchanged, the secondary coil does not generate induced electromotive force, and the current of the secondary coil is zero, so CD is wrong; so select B
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