Variables in physics often include a subscript. What are subscripts used for in physics?
1 answer:
Answer:
C.) To indicate different versions of the same variable.
Explanation:
Variables in physics often include a subscript. These subscripts are used for indicating different versions of the same variable in physics.
Basically, subscripts are used to represent the beginning (initial) and ending (final) position or point of a variable in physics.
For example, we would look at Gay Lussac' Law of gases.
Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Gay Lussac's law is given by;
Where;
represents the initial temperature.
represents the initial temperature.
represents the initial pressure.
represents the initial pressure.
Note: 1 and 2 are the subscript while T and P are the variables.
You might be interested in
Answer:
The net power needed to change the speed of the vehicle is 275,000 W
Explanation:
Given;
mass of the sport vehicle, m = 1600 kg
initial velocity of the vehicle, u = 15 m/s
final velocity of the vehicle, v = 40 m/s
time of motion, t = 4 s
The force needed to change the speed of the sport vehicle;
The net power needed to change the speed of the vehicle is calculated as;
<h2>Answer:</h2>
(a) standard deviation = σ = 4.9996
(b) variance = σ² = 24.996
<h2>Explanation:</h2><h2 /><em>Given frequency table (find attached as Table 1);</em>
<u></u>
(a) To find the sample standard deviation and sample variance, follow these steps;
<em>i. Calculate the mid-point c for each group by using the mid-point formula;</em>
c = (lower bound + upper bound) / 2
=> c = (6.51 + 8.50) / 2 = 7.505
=> c = (8.51 + 10.50) / 2 = 9.505
=> c = (10.51 + 12.50) / 2 = 11.505
=> c = (12.51 + 14.50) / 2 = 13.505
=> c = (14.51 + 16.50) / 2 = 15.505
<em>So the new table becomes (find attached as Table 2);</em>
<em>ii. Calculate the total number of samples (n) which is the sum of all the frequencies.</em>
n = 50+18+42+20+46
n = 176
<em>iii. Calculate the mean (M)</em>
This is done by first multiplying the midpoints by the corresponding frequencies and then dividing the result by the total number of samples (n).
M = [(7.505 x 50) + (9.505 x 18) + (11.505 x 42) + (13.505 x 20) + (15.505 x 46)] / 176
M = [375.25 + 171.09 + 483.21 + 270.1 + 713.23] / 176
M = [2012.88] / 176
M = 11.44
<em>iv. Find the variance (σ²);</em>
The variance is calculated using the following formula
σ² = [Σ(f x c²) - (n x M²)] / (n - 1) ------------(i)
Where;
f = frequency of each boundary data point
<em>=> Let's first calculate </em>Σ(f x c²).
This is done by finding the sum of the product of the frequency (f) of each boundary point and the square of their corresponding mid-points(c)
Σ(f x c²) = [(50 x 7.505²) + (18 x 9.505²) + (42 x 11.505²) + (20 x 13.505²) + (46 x 15.505²)]
Σ(f x c²) = [(2816.25125) + (1626.21045) + (5559.33105) + (3647.7005) + (11058.63115)]
Σ(f x c²) = 24708.1244
<em>=> Now calculate (n x M²)</em>
n x M² = 176 x 11.44²
n x M² = 23033.7536
<em>=> Now substitute these values into equation (i) to calculate the variance</em>
σ² = [Σ(f x c²) - (n x M²)] / (n - 1)
σ² = [24708.1244 - 23033.7536] / (176 - 1)
σ² = [4374.3708] / (175)
σ² = 24.996
Therefore, the variance is 24.996
<em>v. Find the standard deviation (σ)</em>
The standard deviation is the square root of the variance. i.e
σ = √σ²
σ = √24.996
σ = 4.9996
Therefore, the standard deviation is 4.9996