Its (a) and(a)for the other ? and the last one is (d)
If I were to climb a flight of ladfer I would gain potential energy because I would be moving further away from the centre of the earth.
Answer:
![T = 15 \pm 3.89 ^oC](https://tex.z-dn.net/?f=T%20%3D%2015%20%5Cpm%203.89%20%5EoC)
Explanation:
As we know that the equation to convert Fahrenheit to Celcius we have
![C = \frac{5}{9}(F - 32)](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B5%7D%7B9%7D%28F%20-%2032%29)
so we have
![T = 59 ^o F](https://tex.z-dn.net/?f=T%20%3D%2059%20%5Eo%20F)
so we have
![T = \frac{5}{9}(59 - 32)](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B5%7D%7B9%7D%2859%20-%2032%29)
![T = 15^o C](https://tex.z-dn.net/?f=T%20%3D%2015%5Eo%20C)
now the uncertainty in temperature is given as
![\Delta C = \frac{5}{9}(\Delta F)](https://tex.z-dn.net/?f=%5CDelta%20C%20%3D%20%5Cfrac%7B5%7D%7B9%7D%28%5CDelta%20F%29)
so we have
![\Delta T = \frac{5}{9}(7)](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%20%5Cfrac%7B5%7D%7B9%7D%287%29)
![\Delta T = 3.89 ^o C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%203.89%20%5Eo%20C)
so the temperature is given as
![T = 15 \pm 3.89 ^oC](https://tex.z-dn.net/?f=T%20%3D%2015%20%5Cpm%203.89%20%5EoC)
Answer:
497.6 N
Explanation:
From the question,
The net force on the skydiver = weight of the skydiver- the resistive force of air
F' = W-F...................... Equation 1
Where W = weight of the skydiver, F = resistive force of air.
But,
W = mg................ Equation 2
Where m = mass of the skydiver, g = acceleration due to gravity.
Substitute equation 2 into equation 1
F' = mg-F............ Equation 3
Given: m = 87 kg, F = 355 N, g = 9.8 m/s²
Substitute these values into equation 3
F' = 87(9.8)-355
F' = 852.6-355
F' = 487.6 N