Answer:
(a)
(b)
Explanation:
We can derive the initial speed of the rock from the equation of the speed in function of the time:
Using the given values for the speed at time t=1.7s, we get:
In words, the speed of the rock at launch is 34m/s (a).
Next, we use this to calculate the speed at t=4.9s:
This means that the speed of the rock at 4.9s after the launch is 14m/s (b), and the negative sign means that it is moving downwards.
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:
here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
Explanation:
As we know that gravitational field is defined as the force experienced by the satellite per unit of mass
so we will have
now in order to find the acceleration of the satellite we know by Newton's II law
so we will have
so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location