Answer is (c), the latent heat of fusion. That is by definition the heat that 1 kg of a substance must absorb to melt in the vicinity of its melting point.
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:
0.25m²
Explanation:
We know that the summation of forces in the vertical direction is zero
So
PA-mg=0
A=mg/p
So
Substituting
A= 75* 9.8/3*10^-3
=0.25m² which is the total shoe area
a. Independent variable: <em>TIME
</em>
b. Dependent variable: <em>DISTANCE</em>
Answer:
V₁ = 6 V
, V₂ = V₃ = 3 V
Explanation:
To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.
* a serial circuit there is only one path for current
in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term
* Parallel circuit in this there are two or more paths for the current
in this circuit the voltage is constant and the east is divided between each branch
with these principles let's analyze the proposed circuit
The DC battery is in parallel with resistor R1 and the equivalent of the other branch,
as in a parallel circuit the voltage is constant
V₁ = 6 V
in the other branch (23) it forms a series construction, where the current is constant
6 = iR₂ + iR₃
as they indicate that each resistance has the same value
6 = 2 iR
V = V₂ = V₃ = 3 V