A. Average speed is weighted mean (1 × 2 + 2 × 3 + 3 × 5 + 4 × 7 + 3 × 9 + 2 × 12.5)/15 = (2 + 6 + 15 + 28 + 27 + 25)/15 = 103/15 = 6.867 b. RMS is square root of 1/15 times sum of squares of speeds Sum of squares is 4 + 9 + 9 + 25 + 25 + 25 + 49 + 49 + 49 + 49 + 81 + 81 + 81 +156.25 + 156.25 = 848.5
c. RMS speed = √(848.5/15) = 7.521
Most likely the speed is the peak in the speed distribution, which is 7.
The answer for the following problem is mentioned below.
The option for the question is "A" approximately.
- <u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
Explanation:
Given:
Spring constant (k) = 240 N/m
amount of the compression (x) = 0.40 m
To calculate:
Elastic potential energy (E)
We know;
<em>According to the formula;</em>
E =
× k × x × x
<u>E = </u>
<u> × k ×(x)²</u>
where;
E represents the elastic potential energy
K represents the spring constant
x represents amount of the compression in the string
So therefore,
Substituting the values in the above formula;
E =
× 240 × (0.40)²
E =
× 240 × 0.16
E =
× 38.4
E = 19.2 J or approximately 20 J
<u><em>Therefore the elastic potential energy of the string is 20 J.</em></u>
Kinetic energy E = m * v^2
<span>Since the acceleration of both books will be -neglecting air resistance - the same, the kinetic energy will be directly proportional to the mass of the book.</span>
Answer: A. 2.0m above ground
Explanation:
Height of the pitcher from the floor of his mound = 1.8m
Height of the mound above the rest of the field = 0.2m
Since the height difference between the mound and the ground the catcher stands on = 0.2m and the pitcher throws the ball horizontally, then the catcher has to hold his glove at (1.8 + 0.2)m
above ground.
= 2.0m above ground (option A)