Answer:
b. 0.6m/s, 0.7m/s, 0.61m/s, 0.62m/s
Explanation:
Precision of a measurement is the closeness of the experimental values to one another. Hence, experimental measurements are said to be precise if they are close to each other irrespective of how close they are to the accepted value. Precision can be determined by finding the range of each experimental value. The measurement with the LOWEST RANGE represents the MOST PRECISE.
Note: Range is the highest value - lowest value
Set A: 1.5 - 0.8 = 0.7
Set B: 0.7 - 0.6 = 0.1
Set C: 2.4 - 2.0 = 0.4
Set D: 3.1 - 2.9 = 0.2
Set B has the lowest range (0.1), hence, represent the most precise value.
Answer:
Atoms of tellurium (Te) have the greatest average number of neutrons equal to 76.
Explanation:
In the periodic table, Elements are represented with their respected symbols. Above the symbol is the elements atomic number which is equal to the number of protons in each atom. Below the symbol is the mass number of that element which is roughly equal to the sum of neutrons and protons of that atom.
To calculate the number of neutrons we can take the difference of Atomic number and mass number:
Number of neutrons = mass number - atomic number
<u>- Tin:</u>
Atomic number = 50
Mass number = 119
Number of neutrons = mass number - atomic number = 119 - 50
Number of neutrons = 69
<u>- Antimony(Sb):</u>
Atomic number = 51
Mass number = 122
Number of neutrons = mass number - atomic number = 122 - 51
Number of neutrons = 71
<u>- Tellurium(Te):</u>
Atomic number = 52
Mass number = 128
Number of neutrons = mass number - atomic number = 128 - 52
Number of neutrons = <u>76</u>
<u>- Iodine(I):</u>
Atomic number = 53
Mass number = 127
Number of neutrons = mass number - atomic number = 127 - 53
Number of neutrons = 74
Here, the greatest number of neutrons is for the atoms of Tellurium(Te).
Liquid water because if it said very high then it would be water vapor but it didn’t say that so the answer is B liquid water
Answer:

Explanation:
This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the car in this problem,
u = 27.8 m/s
v = 0
s = 17 m
Solving for a, we find the acceleration:
