Answer:
The material cost for making one ton of the brass sample that I have is $8149.04.
Explanation:
Density of copper = 8.96 g/cm^3 = 8.96×10^-3 kg/cm^3
Price of copper = $6.13/kg
Price of copper per volume = 8.96×10^-3 kg/cm^3 × $6.13/kg = $0.0549/cm^3
Density of zinc = 7.14 g/cm^3 = 7.14×10^-3 kg/cm^3
Price of zinc = $1.8/kg
Price of zinc per volume = 7.14×10^-3 kg/cm^3 × $1.8/kg = $0.0129/cm^3
Price of brass per volume = 0.0549 + 0.0129 = $0.0678/cm^3
Density of brass I have is 8.32 g/cm^3 = 8.32 g/cm^3 × 1 kg/1000 g × 1 ton/1000 kg = 8.32×10^-6 ton/cm^3
Price = $0.0678/cm^3 ÷ 8.32×10^-6 ton/cm^3 = $8149.04/ton
We know that a wave is a disturbance that transfers energy through matter or space There are two main types of waves: Mechanical and Electromagnetic. Water waves are mechanical. A mechanical wave is an oscillation of matter to transfers energy, but you always need a medium (substance such as: solid, liquid, gas, plasma) to transport it. The medium for water waves is, in fact, the water. For example, ripple in water is a surface wave. On the other hand, electromagnetic waves don't need a medium to transport, they can do it through the empty space. Then, this is the major characteristic that makes these two types of waves different.
Answer:
This procces is called evaporation.
Explanation:
When you have liquid water that is transformed into steam, a phase change is called evaporation. The temperature for the evaporation of water depends on the pressure, for example for water at atmospheric pressure the temperature of evaporation is equal to 100°C. as the pressure increases are achieved evaporation temperatures higher. When that happens, the phase change temperature of the water is not increasing, as the process that takes place is the transfer of latent heat and applies only to changes of phase, that is to say at atmospheric pressure when it has 100% of the steam this will be at 101°C.
Explanation:
Given that,
Object-to-image distance d= 71 cm
Image distance = 26 cm
We need to calculate the object distance
We need to calculate the focal length
Using formula of lens
put the value into the formula
The focal length of the lens is 35.52.
(B). Given that,
Object distance = 95 cm
Focal length = 29 cm
We need to calculate the distance of the image
Using formula of lens
Put the value in to the formula
We need to calculate the magnification
Using formula of magnification
The magnification is 0.233.
The image is virtual.
Hence, This is the required solution.
