Answer:
The density of the metal is 5200 kg/m³.
Explanation:
Given that,
Weight in air= 0.10400 N
Weight in water = 0.08400 N
We need to calculate the density of metal
Let 
be the density of metal and 
be the density of water is 1000kg/m³.
V is volume of solid.
The weight of metal in air is
.....(I)
The weight of metal in water is
Using buoyancy force
We know that,
....(I)
Put the value of 
in equation (I)
Put the value of Vg in equation (II)
Hence, The density of the metal is 5200 kg/m³.
Regions in the milky way where density waves have caused gas clouds to crash into each other are called clumps.Clumps are molecular clouds (interstellar clouds) with higher density,where lots of dust and gs cores resides. These clouds are the beginning of stars.
When the pump removed the air in the bell, the balloon expanded.
<u>Option: B</u>
<u>Explanation:</u>
In order to construct our own environment in the glass jar known as bell jar system, which can be used to explore and consider our larger environment on Earths, for an instance. Here a glass jar that hinges on an airtight rubber basis i.e seals appropriately. At the top of the jar, a bung is connected to it which passed via a metal tube. It has an adjacent flexible tube that goes to a hand vacuum pump and the best hand-powered pump was made with a wine preserver.
When the pump extracts the air from the bell jar, the pressure inside the balloon naturally decreases. The balloon usually has a air pressure around it, which restricts its size, but when this air is extracted and the pressure around it decreases the gas in the balloon will expand and the balloon seems to be inflating. When you release the air back into the bell jar, it will once again compress back to its actual size.
Answer:
no, when a plastic rod is rubbed with a duster, electrons are transferred from one material to the other. The material that gains electrons becomes negatively charged. The material that loses electrons becomes positively charged.
Explanation:
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
