Swiss astrophysicist Stephanie Udry (right) and University of Geneva astronomer Michel Meyer (left) hold a picture of the planetary system around red dwarf 581 (upper right corner) on April 24, 2007. The planet, code named "581c", is 190 trillion kilometers away from the earth and is orbiting a planet smaller than the sun and hotter than the sun It has been measured that its mass is about 5 times that of the earth and its diameter is about 1.5 times that of the earth A. The ratio of the gravitational acceleration on the surface of planet 581c to the gravitational acceleration on the surface of the earth can be obtained. B. the ratio of the first cosmic velocity between planet 581c and the earth can be obtained. C. The ratio of the circumfluence velocity between planet 581c and the earth can be obtained. D. the ratio of the circumfluence period between planet 581c and the earth can be obtained

Swiss astrophysicist Stephanie Udry (right) and University of Geneva astronomer Michel Meyer (left) hold a picture of the planetary system around red dwarf 581 (upper right corner) on April 24, 2007. The planet, code named "581c", is 190 trillion kilometers away from the earth and is orbiting a planet smaller than the sun and hotter than the sun It has been measured that its mass is about 5 times that of the earth and its diameter is about 1.5 times that of the earth A. The ratio of the gravitational acceleration on the surface of planet 581c to the gravitational acceleration on the surface of the earth can be obtained. B. the ratio of the first cosmic velocity between planet 581c and the earth can be obtained. C. The ratio of the circumfluence velocity between planet 581c and the earth can be obtained. D. the ratio of the circumfluence period between planet 581c and the earth can be obtained

It is known from the title that the mass of the planet code "581c" is 5 times of the mass of the earth, and its diameter is 1.5 times of the earth. The influence of rotation is ignored on the surface of the star. Gmmr2 = mg, i.e. g = gmr2, is known from the universal gravitation equal to gravity. The ratio of the gravitational acceleration on the surface of the planet "581c" to the gravitational acceleration on the surface of the earth can be obtained from the known quantity

When a 2kg mass object is placed on the horizontal ground, u = 0.2, and the horizontal thrust of 10N is applied today, how much displacement will it reach after 5S? That u is Niu,

Because u = 0.2, the friction force F = UMG = 4N is pushed by 10N, so the resultant force is 10-4 = 6N = ma
A = 3m / S2 x = 0.5at * t the displacement = 0.5 * 3 * 25 = 37.5m can be obtained by substituting t = 5S

A physics problem about kinetic energy application in high school
When an object is placed on an inclined plane with a height of 1 m and an inclination of 30 °, it begins to slide. When it moves to the lowest point of the inclined plane, its velocity is v = 4 m / s. The dynamic friction coefficient between the object and the inclined plane is μ

The friction force is μ mgcos30 = √ 3mg μ / 2
Friction work FS = - 2fH = - 3 μ MGH
Gravity work MGH
So the kinetic energy theorem is: MGH - √ 3 μ MGH = 1 / 2 (MV ^ 2)
2(1-√3μ)gh=v^2
μ=√3/15

As shown in the figure, a car is stationary on a horizontal plane, and an object with a mass of M = 8kg is placed on the car. It is pulled and stationary on the car by a spring stretched in the horizontal direction. At this time, the spring force is 6N. Now, a force is applied to the car along the horizontal right direction, so that the car starts to move from stationary, and the acceleration gradually increases from zero to 1m / S2, Then make a uniform acceleration linear motion with the acceleration of 1m / S2
A. When the acceleration of the car is 0.75 & nbsp; m / S2, the object is not affected by friction. C. the object and the car always keep relatively static, and the force of the spring on the object never changes
D. When the car moves in a straight line with a uniform acceleration of L & nbsp; m / S2, the friction force on the object is 8N
A. According to Newton's second law, the resultant force of the block increases from 0 to 8N, the direction of friction decreases to zero to the left, and then increases to the right. In the whole process, the object is stationary relative to the car, and the spring force does not change, A is wrong. & nbsp; & nbsp; & nbsp; B. when the acceleration of the trolley (to the right) is 0.75m/s2, the resultant force of the block f = ma = 6N, and the spring force is equal to 6N, then the friction force is zero. Therefore, B is correct. & nbsp; & nbsp; & nbsp; D. when the trolley moves straight line with uniform acceleration to the right at 1m / S2, the resultant force of the block f = ma = 8N, and the spring force is equal to 6N, then the friction force is 2n, The direction is horizontal to the right. So D is wrong. So BC is selected
What I want to ask is why the spring force does not change

In fact, this question is not very rigorous. It doesn't say whether the spring tension is left or right at the beginning. According to the understanding of the solution, it should be right! Let's look at it from the right understanding. First of all, if you want to make the object move relative to the car, then the resultant force required must be greater than the maximum static friction force. From the beginning 6N, we can see that the maximum static friction force will