The Milky Way galaxy is the one that the sun is a member of, and it contains
our solar system. We're in it, and you can't get much closer than that.
The Milky Way is known to be bigger than your average galaxy, but it's
probably not correct to say that it contains the 'most' stars of any galaxy.
The estimate for the Milky Way is only a few hundred billion stars.
(h + .16) m g = 1/2 k x^2 total PE of block relative to where it stops
(h + .16) .82 * 9.8 = .5 * 120 * .16^2 PE released = PE of spring
8.04 h + 1.29 = 1.536
h = (1.536 - 1.29) / 8.04 = .031 m = 3.1 cm
Answer:
a) > x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
b) 
And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
Explanation:
Part a
For this case we have the following data:
x: 1,2,3,4,5
y: 1.9,3.5,3.7,5.1, 6
For this case we can use the following R code:
> x<-c(1,2,3,4,5)
> y<-c(1.9,3.5,3.7,5.1,6)
> linearmodel<-lm(y~x)
And the output is given by:
> linearmodel
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
1.10 0.98
Part b
For this case we have the following trend equation given:

And if we compare this with the general model 
We see that the slope is m= 0.98 and the intercept b = 1.10
Answer:
60N
Explanation:
in this case the minimum amount of force required must be equal to the friction Force. i.e <u>Newton</u><u>'s</u><u> </u><u>first</u><u> </u><u>law</u><u> of</u><u> </u><u>mot</u><u>ion</u><u>.</u>
therefore the maximum amount of frictional force is equal to the applied force which is 60N.
because of the net force acting on the object is zero the object is in constant motion . i.e equal and opposite force must be applied so that the object is in constant velocity therefore the total frictional force must be 60N
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object



v=5.02 m/s
c.Work done by friction on the incline,
