To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by
Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that
Therefore,
Re-arrange to find x,
The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.
I think it is D because u think of space and Electromagnetic Waves Travel Through Space Moving All Kinds of Particles From The Sun With It Which Creates The Northern and Southern Lights In The Poles :-)
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 peak emf of the generator is 40.94 V.
Explanation:
Given that,
Number of turns in primary coil= 11
Number of turns in secondary coil= 18
Peak voltage = 67 V
We nee to calculate the peak emf
Using relation of number of turns and emf
Where, N₁ = Number of turns in primary coil
N₂ = Number of turns in secondary coil
E₂ = emf across secondary coil
Put the value into the formula
Hence, The peak emf of the generator is 40.94 V.
