How many times does 1.502 expand or shrink without the decimal point?

Removing the decimal point is equivalent to moving the decimal point three places to the right, so it is expanded 1000 times
Move right to enlarge, move left to shrink
Move one bit 10 times, two bits 100 times, three bits 1000 times

After the decimal point of 0.45 is removed, it is expanded to several times of the original

A hundred times

Km / h equals__________ m/s

1m/s=3.6km/h
So 1km / h = (5 / 18) m / S
If you don't understand, please ask, I wish you a happy study!

On a 1:2000000 scale map, 2cm represents the actual distance () km

On a 1:2000000 scale map, 2 cm represents the actual distance of (40) km

On a map with a scale of 1:300000, the distance between a and B is 2.4cm
What is the distance between a and B on the map?

I'm a sixth grade student. I've done this problem. I'll do it for you
Actual distance = distance on the map divided by scale, then:
2.4 / (1 / 300000)
=2.4*300000
=720000 (CM)
Distance on graph = actual distance multiplied by scale
720000 * (1 / 200000)
=3.6 (CM)
In fact, this question is very simple. I hope you can adopt it

On a map with a scale of 1:300000 and a scale of 1:200000, the distance between a and B is 10.2 cm. If you draw a map of a and B, how many cm is the distance between a and B?

The actual distance is 10.2 × 300000 = 3060000 (CM) when the scale is 1:300000
If the map is redrawn on the map of 1; 200000, then: 3060000 △ 200000 = 15.3 (CM)
A: if it is redrawn on a map with a scale of 1; 200000, the distance between a and B is 15.3cm

On a map with a scale of 1:300000, the distance between a and B is 6cm. On another map with a scale of 1:200000, what is the distance between the two places?

A: the distance between the two places on the map is 9cm

On a map with a scale of 1:300000, the distance between a and B is 6cm. On a map with a scale of 1:200000, what is the distance between a and B?

Distance = 6 × 300000 △ 200000 = 9cm

The plane flies in the direction of the earth's rotation and in reverse
Why do you say "when the plane and the earth rotate in the same direction, it means that the plane does not move? People on the plane will see sunrise and sunset twice a day 24 hours a day? When flying in the opposite direction, if it takes off in the daytime, it will always be daytime?"
Hope to explain clearly!

To explain this problem, let's first explain several directions. If we look down at the earth from the north pole, then the earth's rotation direction is counterclockwise. Now suppose the earth is stationary, then the sun moves around the earth clockwise. At the same time, if the aircraft flies along the earth's rotation direction, it flies around the earth counterclockwise. On the contrary, if it flies against the earth's rotation direction, it flies clockwise
For the first problem, as the reference of the aircraft and the sun, we assume that the earth is stationary, and the aircraft flies along the direction of the earth's rotation, that is, the flight direction of the aircraft is counter clockwise (still the north pole direction overlooking the earth), and the sun's motion direction is clockwise, within the time that the sun rotates around the earth for one cycle (24 hours), The plane and the sun will meet twice (assuming that it takes 24 hours for the plane to circle around the earth, and then explain the influence of speed change). This is what you call "people on the plane will see sunrise and sunset twice 24 hours a day"
The second problem is still based on the previous assumption and observation angle. At this time, the aircraft flies counter to the earth's rotation direction, that is, it flies clockwise. At the same time, the sun also moves clockwise around the earth. In this case, it is equivalent to the aircraft and the sun moving in the same direction (also assuming that it takes 24 hours for the aircraft to revolve around the earth). Due to the synchronization of the sun and the aircraft, there will be "counter direction flight", If you take off in the daytime, it will always be daytime
But in real life, the speed of an ordinary airliner is about 900 km / h, and it takes 40000 / 900 = 44 hours to circle the equator, which is not as fast as the sun (or not as fast as the earth's rotation). This will make the above phenomenon hold in some cases (mainly depending on the relative position of the plane and the sun at the time of departure, that is, the time in real life, It may take the longest time for a plane to fly around the equator. The closer to the high latitude, the shorter the time it takes for a plane to fly around the earth, which is closer to the above assumption (it takes 24 hours for a plane to fly around the equator), and the more likely the phenomenon mentioned above will appear
Personal understanding, for your discussion

The mass of the earth is about 81 times that of the moon. An aircraft is between the earth and the moon. When the gravity of the earth is equal to that of the moon, the ratio of the distance between the aircraft and the center of the earth to the center of the moon is ()
A. 27：1B. 1：27C. 1：9D. 9：1

If the mass of the moon is m, the mass of the earth is 81m. The distance between the spacecraft and the center of the earth is R1, and the distance between the spacecraft and the center of the moon is R2. Because the gravity of the earth and the moon is equal, according to the law of universal gravitation, we can get the following result: g81mmr21 = gmmr22, and the solution: r1r2 = 811 = 91. So option D is correct

How many times does 1.502 expand or shrink without the decimal point?