Directly proportional to pressure
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
Answer:
When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.
Hope it helps u:)
Answer:
ReCrystallization
Contact pressure causing grains to "grow" together.
It's a metamorphic process used to rearrange the atoms of the minerals in order to modify them in to the required form under the specific temperature and pressure. For example conversion of limestone into marble.
<em>Cementation:</em>
Precipitation of bonding agents between grains.
Cementation is the process in which the minerals in the form of fluid fills in spaces between the grains and binds them together upon crystallizing.
<em>Compaction:</em>
Increase in density due to weight of overburden.
Compaction as it's name indicate is the compaction of the sediments due to the heavy weight of the rocks which reduce their size and increase their density.
Answer: Electric field vector is created by an electric force of a charged particle.
Explanation: The direction of the field vector can be determined by applying Coulomb’s Law which explains the electric force between charge particles. If q is positive the force is repulsive, to a test charge ( which is positive always ) and if q is negative the force is attractive to a test charge.
