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:
The correct question is:
"Find the energy each gains"
The energy gained by a charged particle accelerated through a potential difference is given by

where
q is the charge of the particle
is the potential difference
For a proton,

And since 
The energy gained by the proton is

For an alpha particle,

Therefore, the energy gained is

Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)

So the energy gained is the same as the proton:

Answer:
120 m/s
Explanation:
Given:
v₀ = 0 m/s
a = 12 m/s²
t = 10 s
Find: v
v = at + v₀
v = (12 m/s²) (10 s) + 0 m/s
v = 120 m/s
Because they behave just like all the electromagnetic waves of the spectrum. Same equations, just shorter wavelengths and more energy.
Hope you get it :)