Which statement best describes air pressure at high altitudes? (Points : 3)
2 answers:
You might be interested in
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:
t = 0.714 s and x = 5.0 m
Explanation:
This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed
vₓ = 7.0 m / s
Let's find the time it takes to get to the river
y = y₀ + v_{oy} t - ½ g t²
the initial vertical speed is zero and when it reaches the river its height is zero
0 = y₀ + 0 - ½ g t²
t =
t = ra 2 2.5 / 9.8
t = 0.714 s
the distance traveled is
x = vₓ t
x = 7.0 0.714
x = 5.0 m