Answer:
a) x_average = ∑ 
/ n
, b) Δx_{i} = x_{i} –x_average,
d) σ = √(1/n-1 ∑ Dx_{i}² )
Explanation:
Some definitions are requested
a) the average value is the sum of all the values divided by the number of them, if the uncertainties are random, this is the closest value to the real one
x_average = ∑ / n
b) The deviation from the mean value or absolute error is the measured value minus the average value
Δx_{i} = x_{i} –x_average
c) is the average value of the deviations
Δx_average = ∑ Δx_{i} / n
d) It is a measure of the dispersion of the values with respect to their average value, it takes the worst of all cases, widely used for large numbers of data
σ = √(1/n-1 ∑ Dx_{i}² )
Experimental results should be given as follows
Average value ± uncertainty and the standard deviation
(x_average + - Δx_average)
σ
(c) as the change in the dependent variable is in direct CORRELATION to the change in the independent variable.
A run though an open field during a thunderstorm is the answer
Answer:
40N
Explanation:
Since both weights are connected to one string, you can say that the tensions above each are equal to each other.
If you do the sum of forces for the 4kg mass, then the tension comes out to 40N (if we take gravity to be 10m/s²). But that seemed too good to be true, so I decided to do the work for the 7kg mass as well [which included finding the normal force (N) and plugging it into the sum of forces for the 7kg mass] to find that it also gives 40N as the answer.
If I were to put my process into steps:
- Write out the sum of Forces for both masses
- Set them equal to each other to find normal force (because this is the only unknown)
- Calculate and compare the two tensions to see if they are equal
*This all seems to line up perfectly, but do let me know if my answer doesn't match up with what you might find to he the answer later on.
Answer:
top speed = 17.25
Total height = 281.19 m
Explanation:
given data
mass = 75 kg
thrust = 160 N
coefficient of kinetic friction = 0.1
solution
we get here frictional force acting that is
frictional force = 
.............1
frictional force = 0.1 × 75 × 9.8
frictional force = 73.5 N
and
Net force acting will be F = 160 - 73.5 N
F = 86.5 N
so
Acceleration in the First 15 second will be
F = ma .........2
86.5 = 75 × a
a = 1.15 m/s²
and
now After 15 second the velocity will be as
v = u + at ..........3
here u is 0
so v will be
V = 1.15 × 15
v = 17.25
and
now we get travels distance S in 15 s
s = u × t + 0.5 × a × t²
here u is 0
so distance s will be
s = 0.5 × a × t²
s = 0.5 × 1.15 × 15²
s = 129.37 m
and
now acceleration acting is
F =
m a =
a = 
a = - 0.98
here it is negative it mean downward nature of acceleration
and
now we get distance s by this formula
V² - u² = 2 a s
here v velocity is 0 and
u initial velocity is 17.25 m/s
put here value
0 - 17.25² = 2 × (-0.98) × s
solve it we get
s = 151.82 m
so
Total height is
Total height = 129.37 m + 151.82 m
Total height = 281.19 m
