Answer with Explanation:
We are given that
Area of loop=

Resistance, R=
B=
We know that magnetic flux

Emf ,
Current, 
Current, 
Substitute t=0 s
Then, I=
=1.6 A
Substitute t=1 s
Then, I=
=0
Substitute
t=2 s
Current, I=
=1.6 A
Answer:
Explanation:
Density = Mass / Volume = 850 / 40*10*5 = 0.425 g /cm^3
Answer:
The only such elements are the Noble Gases (He, Ne, Ar, Kr, Xe, Rn)
(that is helium, neon, argon, krypton, xenon and radon)
Term: Monoatomic
Explanation:
There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.
What is the minimum runway length that will serve?
Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.
A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.
Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.
To learn more about runway
brainly.com/question/11553726
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Answer:
A. h = 2.15 m
B.
Pb' = 122 KPa
Explanation:
The computation is shown below:
a) Let us assume the depth be h
As we know that

After solving this,
h = 2.15 m
Therefore the depth of the fluid is 2.15 m
b)
Given that
height of the extra fluid is

h' = 0.355 m
Now let us assume the pressure at the bottom is Pb'
so, the equation would be

Pb' = 122 KPa