1.A 0.25g flea sits on a disc at a distance of 4cm from the center of the disc.if the disc rotates at 67 rpm and the flea is just able to maintain its position without sliding,determine the coefficient of static friction between the flea and the disc

1.A 0.25g flea sits on a disc at a distance of 4cm from the center of the disc.if the disc rotates at 67 rpm and the flea is just able to maintain its position without sliding,determine the coefficient of static friction between the flea and the disc
1.A 0.25g flea sits on a disc at a distance of 4cm from the center of the disc.if the disc rotates at 67 rpm and the flea is just able to maintain its position without sliding,determine the coefficient of static friction between the flea and the disc that enables the flea to maintain its uniform circular motion.
2.Snoopy is flying his vintage war plane in loop the loop path chasing the red Baron.his instruments tells him that the plane is level (at the bottom of the loop) and travelling at a speed of 180km/h.he is sitting on a bathroom scale,and notes that it reads four times the normal force of gravity on him.what is the radius of the loop?
3.As an indication of the size of the sun's gravitational pull on earth,carry out the rough calculations that follow.suppose that the sun's gravitational attraction could be replaced by a steel wire,running from the sun to the earth,with the wire's tension holding the earth in its orbit.good steel has a breaking stress of 5.0x10^8N/m^2 of cross section area.
a)calculate the cross-section area of wire that could just hold the earth in its orbit.
b)calculate the corresponding wire diameter.
4.By looking at distance galaxies,astronnomers have conclude that our solar system is circling the center of our galaxie.the hub of this galaxie is lacated about 2.7x10^20m from our sun,and our sun circles the center about every 200 million years.we assume that our sun is attracted by a large number of stars at the hub of the galaxy,and that the sun is kept in orbit by the garavitational attraction of these stars.
a)calculate the total masss of the star at the hub of our galaxy
b)based on an average size star of mass 2.0x10^30kg,determine the approximate number of such stars at the hub
请把公示写清楚点 不要只给我答案
可以用以下公示：
Fnet=mxa
Fc=mxac
Fg=mg
Fg=Gm1m2/d^2
脑子不太好使 呵呵 帮下忙

1.A 0.25g flea sits on a disc at a distance of 4cm from the center of the disc.if the disc rotates at 67 rpm and the flea is just able to maintain its position without sliding,determine the coefficient of static friction between the flea and the disc
1.A 0.25g flea sits on a disc at a distance of 4cm from the center of the disc.if the disc rotates at 67 rpm and the flea is just able to maintain its position without sliding,determine the coefficient of static friction between the flea and the disc that enables the flea to maintain its uniform circular motion.
A) calculate the force applied to the flea
F=ma=mω^2*r=m*r*(2π/T)^2,where T=1/n,n is the rpm (i.e.67)
B) then calculate the static coefficient of the flea.
The formula for the static coefficient is given by F(in Newtons) =mgx
where m=mass in kg,g = 9.81 & x is the static coefficient
2.楼上已经解释的很清楚了
3.As an indication of the size of the sun's gravitational pull on earth,carry out the rough calculations that follow.suppose that the sun's gravitational attraction could be replaced by a steel wire,running from the sun to the earth,with the wire's tension holding the earth in its orbit.good steel has a breaking stress of 5.0x10^8N/m^2 of cross section area.
a)calculate the cross-section area of wire that could just hold the earth in its orbit.
b)calculate the corresponding wire diameter.
A) Fg=Gm1m2/d^2
which is the force that the wire must hold.So G(of sun)*mass of sun*Mass of Earth* / the distance squared.
(this is the equation you gave).
That force divided by the breaking stress per area (5.0x10^8) is equal to the area of the wire.
B) Area = pi*r^2.& diameter = 2r.simple enough.
4.By looking at distance galaxies,astronomers have conclude that our solar system is circling the center of our galaxy.the hub of this galaxy is located about 2.7x10^20m from our sun,and our sun circles the center about every 200 million years.we assume that our sun is attracted by a large number of stars at the hub of the galaxy,and that the sun is kept in orbit by the gravitational attraction of these stars.
a)calculate the total mass of the star at the hub of our galaxy
Fg=Gm1m2/d^2
Assume that the sun follows a circular pathway,the distance the sun travels is calculated with the radius given (2.7x10^20m).And the speed can be calculated.Knowing that while in orbit,a satellite is always accelerating.F=ma=mω^2*r=m*r*(2π/T)^2,hence,you have the force and acceleration.
Next,you can calculate m1 by substituting in for all values.
b)based on an average size star of mass 2.0x10^30kg,determine the approximate number of such stars at the hub
m1/mass per star.
Message me if you don't understand the solution.How I miss high school physics.^__^

我向你发誓：这些题都不难。
我同样向你发誓：这些英语真的很烦……
你是不是在努力出国？我不知道外国的公式，只会用中国的。
1 运用匀速圆周运动的公式F=ma=mω^2*r=m*r*(2π/T)^2,而又因为n（转速）=1/T,所以就是F=m*r*(2π*n)^2,其中都要化成国际单位，然后代入运算。
2 这道题我没全看懂，大概意思可能看懂了。它说自己看到了四倍的体重...

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我向你发誓：这些题都不难。
我同样向你发誓：这些英语真的很烦……
你是不是在努力出国？我不知道外国的公式，只会用中国的。
1 运用匀速圆周运动的公式F=ma=mω^2*r=m*r*(2π/T)^2,而又因为n（转速）=1/T,所以就是F=m*r*(2π*n)^2,其中都要化成国际单位，然后代入运算。
2 这道题我没全看懂，大概意思可能看懂了。它说自己看到了四倍的体重，那么就是当时在圆周运动底部的时候飞机给了他（它）四倍体重的支持力，而减去一倍体重，得到向上三倍体重的向心力。F（向心）=3mg，然后除以m，得到向心加速度a=3mg,有因为a=v^2/r,两个式子联立就可以得到r。
3 估计是把引力算出来然后看看多粗的绳子可以代替。
4 就是套公式，你随便找本《新专题教程》（力学）都有。

很抱歉，最后两道实在是看不懂了。

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