When a boy throws a ball and accidentally breaks a window, the momentum of the ball and all the pieces of glass taken together after the collision is THE SAME as the momentum of the ball before the collision
hope this helps
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
If the force were constant or increasing, we could guess that the speed of the sardines is increasing. Since the force is decreasing but staying in contact with the can, we know that the can is slowing down, so there must be friction involved.
Work is the integral of (force x distance) over the distance, which is just the area under the distance/force graph.
The integral of exp(-8x) dx that we need is (-1/8)exp(-8x) evaluated from 0.47 to 1.20 .
I get 0.00291 of a Joule ... seems like a very suspicious solution, but for an exponential integral at a cost of 5 measly points, what can you expect.
On the other hand, it's not really too unreasonable. The force is only 0.023 Newton at the beginning, and 0.000067 newton at the end, and the distance is only about 0.7 meter, so there certainly isn't a lot of work going on.
The main question we're left with after all of this is: Why sardines ? ?
Horizontal distance covered by a projectile is X = Vix *T
where Vix is the initial horizontal component of velocity and T is time taken by the projectile
Vix = ViCos theta
In question they said that initial velocity and angle is same on earth and moon
so Vix would remains same
now let's see about time taken T
time taken to reach the highest point
Vfy = Viy +gt
at highest point vertical velocity become zero so Vfy =0
0 = Vi Sin theta + gt
t = Vi Sintheta /g
Total time taken to land will be twice of that
On earth
Te= 2t
Te = 2Sinθ/g
on moon g is one-sixth of g(earth)
Tm = 2Sinθ/(g/6)
Tm = 6(2Sinθ/g)
Tm = 6Te
so total time taken by the projectile on moon will be six times the time taken on earth
From first equation X = Vix*T
we can see that X will also be 6 times on moon than earth
so projectile will cover 6 times distance on moon than on earth
Explanation:
Given that,
Length of gold wire, l = 4 m
Voltage of battery, V = 1.5 V
Current, I = 4 mA
The resistivity of gold, 
Resistance in terms of resistivity is given by :
Also, V = IR
So,
A is area of wire,
, r is radius, r = d/2 (diameter=d)
Out of four option, near option is (C) 17 μm.
