Answer:The coefficient of friction between the box and the floor, = 1.456 × 10⁻²
Explanation:
B. Earth’s outer surface is cooler than its interior layers.
Explanation:
- The option given above is showing us that the temperature in the interior of the earth is higher than the temperature in the outer layer.
- There is travel of heat from the inner core of the earth to the earth's crust. Due to the loss of heat when it reaches the outer layer, there arises a temperature difference.
- The heat loss is due to the absorption of heat during its transfer. Hence, option B is the answer.
Answer: 0.0138 m^2 = 138 cm^2
Explanation:
The thermal expansion is the term use for the physical phenomena of dilation of the objects when they are exposed to changes in temperature.
The objects dilate when they are heated and contract when they are cooled.
The dilation is proportional to the change in temperatur.
For linear dilation, the proportionality constant is called linear dilation coefficient of the materials, it is named α and is measured in °C ^-1.
ΔL = α * Lo * ΔT, which means that the dilation (or contraction) is proportional to the product of the original length (Lo) and the change of temperature (ΔT).
There is also superficial dilation, for which the dilation is:
ΔA = β * Ao * ΔT, which means that the superficial dilation (or contraction) is proportional to the product of the original area (Ao) and the change of temperature (ΔT).
It is very interesting and important to solve problems that β = 2α, because regularly you will find the values of α for different materials and so, you just to multiply it times 2 to use β.
For this problem:
- Original area, Ao = area of the flat roof at - 10°C = 2.0m * 3.0m = 6.0 m^2.
- α for aluminum = 24 * 10^ -6 °C^-1.
- ΔT = 38°C - (-10°C) = 48°C
So, ΔA = 6.0m^2 * (2 * 24*10^-6 °C&-1) * 48°C = 0.0138 m^2
And that is the area that should stick out in summer to fit the structure during cold winter nights.
You can pass that number to cm^2 to grasp better the idea of this size:
0.0138 m^2 * (100 cm)^2 / m^2 = 138 cm^2
The concepts necessary to solve this problem are framed in the expression of string vibration frequency as well as the expression of the number of beats per second conditioned at two frequencies.
Mathematically, the frequency of the vibration of a string can be expressed as
Where,
L = Vibrating length string
T = Tension in the string
Linear mass density
At the same time we have the expression for the number of beats described as
Where
= First frequency
= Second frequency
From the previously given data we can directly observe that the frequency is directly proportional to the root of the mechanical Tension:
If we analyze carefully we can realize that when there is an increase in the frequency ratio on the tight string it increases. Therefore, the beats will be constituted under two waves; one from the first string and the second as a residue of the tight wave, as well
Replacing 
for n and 202Hz for 
The frequency of the tightened is 205Hz
Answer:
The answer depends on what object you are dropping. Are you dropping a balloon or a car? (I'm joking 'bout that one.) If the mass of the object is very little, then it might drop slower. If the mass is bigger, then it might drop faster.
Good luck!
Explanation:
