Answer:
95 %
99.7 %
Explanation:
= 166 cm = Mean
= 5 cm = Standard deviation
a) 156 cm and 176 cm


From the empirical rule 95% of all values are within 2 standard deviation of the mean, so about 95% of men are between 156 cm and 176 cm.
b) 151 cm and 181 cm


The empirical rule tells us that about 99.7% of all values are within 3 standard deviations of the mean, so about 99.7% of men are between 151 cm and 181 cm.
If the net force on object A is 5 N and the net force on object B is 10 N, then object B will accelerate more quickly than object A provided the mass of both objects are same.
Answer: Option C
<u>Explanation:
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According to Newton’s second law of motion, any external force applied on an object is directly proportional to the mass and acceleration of the object. In order to state this law in terms of acceleration, it is stated that acceleration exhibited by any object is directly proportional to the net force applied on the object and inversely proportional to the mass of the object as shown below:

So if two objects A and B are identical which means they have same mass, then the acceleration attained by the object will be directly proportionate to the net forces exerted on the objects only.
Thus if the force applied is more for one object, then the object will be exhibiting more acceleration compared to the other one. So as object B is experiencing a net force of 10 N which is greater than the net force experiences by object A, then the object B will be accelerating more quickly compared to the object A's acceleration.
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Answer:
1. The period is 1.74 s.
2. The frequency is 0.57 Hz
Explanation:
1. Determination of the the period.
Spring constant (K) = 30 N/m
Mass (m) = 2.3 Kg
Pi (π) = 3.14
Period (T) =?
The period of the vibration can be obtained as follow:
T = 2π√(m/K)
T = 2 × 3.14 × √(2.3 / 30)
T = 6.28 × √(2.3 / 30)
T = 1.74 s
Thus, the period of the vibration is 1.74 s.
2. Determination of the frequency.
Period (T) = 1.74 s
Frequency (f) =?
The frequency of the vibration can be obtained as follow:
f = 1/T
f = 1/1.74
f = 0.57 Hz
Thus, the frequency of the vibration is 0.57 Hz