Answer:
c. 1.5 s
Explanation:
i had this and i got it right
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
Answer:
mass of the neutron star =3.45185×10^26 Kg
Explanation:
When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.
That is

M_ns = mass odf the netron star.
G= gravitational constant = 6.67×10^{-11}
R= radius of the star = 18×10^3 m
ω = 10 rev/sec = 20π rads/sec
therefore,

= 3.45185... E26 Kg
= 3.45185×10^26 Kg
The density of the object is approximately 1.91 kg per m³.
42 kg is a measure of mass, and 22 m³ is a measure of volume. Knowing this, you can use the relationship

to solve for the object's density.
42 kg

22 m³

1.91 kg per m³.
Answer:

Explanation:
If we express all of the cordinates in their rectangular form we get:
A = (1404.77 , 655.06) m


Since we need C to be (0,0) we stablish that:

That way we make an equation system from both X and Y coordinates:


Replacing values:
With this system we can solve for both Db and Dc and get the answers to the question:

