Answer:
0.01
Explanation:
Given the data:
10.1,9.87, 9.76, 9.91, 9.75, 9.88, 9.69, 9.83, 9.90
True value = 9.81
Mean value :
Σx / n
Sample size, n = 9
(10.1 + 9.87 + 9.76 + 9.91 + 9.75 + 9.88 + 9.69 + 9.83 + 9.90) / 9
= 88.69 / 9
= 9.854
Standard deviation (σ) :
Sqrt (Σ(X - m)² / n)
[(10.1 - 9.854)^2 + (9.87 - 9.854)^2 + (9.76 - 9.854)^2 + (9.91 - 9.854)^2 + (9.75 - 9.854)^2 + (9.88 - 9.854)^2 + (9.69 - 9.854)^2 + (9.83 - 9.854)^2 + (9.90 - 9.854)^2] / 9
Sqrt(0.113824 / 9)
Sqrt(0.0126471)
σ = 0.1124593
Standard Error = σ / sqrt(n)
Standard Error = 0.1124593 / 9
Standard Error = 0.0124954
Standard Error = 0.01 ( 1 significant digit)
The best answer would be the 4th choice. "They help scientists explain concepts that are difficult to observe, this also covers the first answer which helps the scientist to answer complex questions. A scientific model is not used prove scientific laws as they may not always have all the data to prove so, instead it is used to allow them to explain better concepts revolving around science through research and may also allow them to predict results based on the accumulation of data and analyzing the trend of this resulting information.
Answer:
a. 4 m/s b. 0.2 V
Explanation:
a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.
The hall voltage V = vBd where v = flow-rate, B = magnetic field strength = 0.500 T and d = diameter of pipe = 3.00 cm = 0.03 m
Since V = vBd
v = V/Bd given that V = 60.0 mV = 0.060 V, substituting the values of the other variables, we have
v = 0.060 V/(0.500 T × 0.03 m)
v = 0.060 V/(0.015 Tm)
v = 4 m/s
b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?
Since the hall voltage, V = vBd and v = flow-rate = 4 m/s, B = magnetic field strength = 0.500 T and d' = diameter of pipe = 10.0 cm = 0.10 m
Substituting the variables into the equation, we have
V = vBd
V = 4 m/s × 0.500 T × 0.10 m
V = 0.2 V
Answer: Smaller than ; larger than
Explanation:
When the elevator is moving in the upward direction, then the force acting on it is negative in nature because of
N= mg +ma, (g is gravity and a is acceleration)
here ma is negative so the N= mg-ma
Hence, it feels smaller than its original weight.
When the elevator is moving downward , then the force acting will be positive in nature
N= mg+ma,
here ma will be positive so it feels larger the original weight of passenger.