A straight line with a negative slope on a velocity-time graph indicates which of the following?
1 answer:
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:
Wavelength = 736.67 nm
Explanation:
Given
Energy of the photon = 2.70 × 10⁻¹⁹ J
Considering:
where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Frequency is:
Frequency = c / Wavelength
So, Formula for energy:
Energy = 2.70 × 10⁻¹⁹ J
c = 3×10⁸ m/s
h = 6.63 x 10⁻³⁴ J.s
Thus, applying in the formula:
Wavelength = 736.67 × 10⁻⁹ m
1 nm = 10⁻⁹ m
So,
<u>Wavelength = 736.67 nm</u>