Solution:
Let the slope of the best fit line be represented by '
'
and the slope of the worst fit line be represented by '
'
Given that:
= 1.35 m/s
= 1.29 m/s
Then the uncertainity in the slope of the line is given by the formula:
(1)
Substituting values in eqn (1), we get
= 0.03 m/s
Answer:
195 is it current pls tell
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
Answer:
Explanation:
Let the potential difference between the plate is V . Then in the first case
Electric field E between plate
E₁ = V / d
where d is separation between plate
When the plate separation becomes d / 2
Electric field E between plate
E₂ = V / d /2
= 2 V / d =2E₁
Or twice the earlier field
Answer:
114.32195122 but Round your answer to three significant figures.) is 114
Explanation:
Just took the test
