Answer:
Explanation:
Descriptive statistics statistics are measures that summarize the characteristics of a sample.
Hope it helps
Answer:
B) The tree was stationary and began to move.
Explanation:
This situation can be explained by using Newton's first law of motion, which states that
"An object at rest (or in motion at constant velocity) stays at rest (or in motion at constant velocity) unless a net non-zero force is exerted on it"
This means that an object at rest can start moving if and only if there is a net non-zero force acting on it.
In the example in the problem, the tree is initially stationary. Later, it started to move. According to Newton's first law, therefore, there must be a net force that caused this change of state of motion of the tree. Therefore, the correct answer is
B) The tree was stationary and began to move.
Answer:
V = 6.65 [volt]
Explanation:
First, we must calculate the power by means of the following equation, where the voltage is related to the energy produced or consumed in a given time.
![P=E/t\\P = 40/30\\P = 1.33[s]](https://tex.z-dn.net/?f=P%3DE%2Ft%5C%5CP%20%3D%2040%2F30%5C%5CP%20%3D%201.33%5Bs%5D)
Using the power we can calculate the voltage, by means of the following equation that relates the voltage to the current.

where:
V = voltage [Volts]
I = current = 200 [mA] = 0.2 [A]
![V = 1.33/0.2\\V = 6.65 [volt]](https://tex.z-dn.net/?f=V%20%3D%201.33%2F0.2%5C%5CV%20%3D%206.65%20%5Bvolt%5D)
Answer:
Mass = 52.63 kilograms
Explanation:
Given the following data;
Acceleration due to gravity = 3.8 m/s²
Weight = 200 N
To find mass;
Mathematically, the weight of a physical object is given by the formula;
Weight = mass * acceleration due to gravity
Substituting into the formula, we have;
200 = mass * 3.8
Mass = 200/3.8
Mass = 52.63 kg
Answer:
21544 N
Explanation:
Let the atmospheric pressure be 101325 Pa
The side length of the cubic box:

![d = \sqrt[3]{V} = \sqrt[3]{0.051} = 0.371 m](https://tex.z-dn.net/?f=d%20%3D%20%5Csqrt%5B3%5D%7BV%7D%20%3D%20%5Csqrt%5B3%5D%7B0.051%7D%20%3D%200.371%20m)
The area of the cubic box:

20 C = 20 + 273 = 293 K
180 C = 180 + 273 = 453 K
As the volume of the air inside the closed cube is not changed, assume the ideal gas law we have

Where P1 = 101325 Pa and T1 = 293K are the original atmospheric pressure and temperature. P2 and T2 = 453 are the new pressure and temperature after the cube gets heat up

The net force on each side of the box it its pressure times side area
