Answer:
<em>A. Statistics addresses gaps in knowledge.</em>
Explanation:
The following statements that does not describe a limitation of statistics is <em>statistics addresses gaps in knowledge.</em>
Answer:
One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.
Explanation:
This hypothesis is based on the fact that the speed of sound in air is v = 343 m / s with a small variation with temperature.
The speed of sound in solid soil is an average of the speed of its constituent media, giving values between
wood 3900 m / s
concrete 4000 m / s
fabrics 1540 m / s
earth 5000 m / s wave S
ground 7000 m / s P wave
we can see that the speed on solid earth is an order of magnitude greater than in air.
One way to test the hypothesis is to create two waves, one in the air and one on the ground at the same time. One of them for the elephant to get closer and another for the elephants to move away. Observe the reaction of the animal and with this we know which sound came first.
From the initial information, the wave going through the ground should arrive first.
The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
<h3>What is the specific heat capacity?</h3>
The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."
Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.
Given data;
Mass of liquid sample of Alcohol m₁ = 200-gram
The temperature of alcohol, T₁ = -6°C.
Mass of liquid sample of water m₂ = 400-gram
The temperature of the water, T₂= 20°C.
The specific heat capacity of the alcohol, S₁=?
The specific heat capacity of water is, S₂=4.19 kJ/kg.°C
As we know that;
<h3 />
![\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C](https://tex.z-dn.net/?f=%5Crm%20Q_%7Bgain%7D%3D%20Q%7Bloss%7D%20%5C%5C%5C%5C%20Q_%7Balcohol%7D%20%3DQ_%7Bwater%7D%20%5C%5C%5C%5C%5C%20m_1s_1%5Ctriangle%20T_1%20%3D%20m_2S_2%20%5Ctriangle%20S_2%20%5C%5C%5C%5C%20200%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20S_1%20%5B%20%2812-%28-6%29%20%5D%20%3D%2040%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%204.19%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%20%2820-12%29%5C%5C%5C%5CS_1%20%3D%202%20%5Ctimes%204.19%20%5Ctimes%2010%5E3%20%5Ctimes%20%5Cfrac%20%7B8%7D%7B18%7D%20%5C%5C%5C%5C%20S_1%20%3D%203.72%20%20%5C%20kJ%20%2Fkg%20%5E0%20C)
Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.
To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.
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Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
Pressure is the force per unit area. This means that the pressure a solid object exerts on another solid surface is its weight in newton’s divided by its area in square metres
