All magnetic fields are created by moving charged particles. So guessed Particles produces magnetic field (electric Currents)
Answer:
1010713.18851 Pa
387999.259089 N
![2.48335668\times 10^9\ Pa](https://tex.z-dn.net/?f=2.48335668%5Ctimes%2010%5E9%5C%20Pa)
Explanation:
= Atmospheric pressure = ![8.043\times 10^4\ Pa](https://tex.z-dn.net/?f=8.043%5Ctimes%2010%5E4%5C%20Pa)
r = Radius = 2 m
h = Depth = 10 m
= Density of liquid methane = ![415\ kg/m^3](https://tex.z-dn.net/?f=415%5C%20kg%2Fm%5E3)
g = Acceleration due to gravity = 7.44 m/s²
A = Area
Force is given by
![F=P_0A\\\Rightarrow F=8.043\times 10^4\times \pi\times 2^2\\\Rightarrow F=1010713.18851\ N](https://tex.z-dn.net/?f=F%3DP_0A%5C%5C%5CRightarrow%20F%3D8.043%5Ctimes%2010%5E4%5Ctimes%20%5Cpi%5Ctimes%202%5E2%5C%5C%5CRightarrow%20F%3D1010713.18851%5C%20N)
The force exerted is 1010713.18851 Pa
![F=mg\\\Rightarrow F=\rho Vg\\\Rightarrow F=415\times \pi 2^2\times 10\times 7.44\\\Rightarrow F=387999.259089\ N](https://tex.z-dn.net/?f=F%3Dmg%5C%5C%5CRightarrow%20F%3D%5Crho%20Vg%5C%5C%5CRightarrow%20F%3D415%5Ctimes%20%5Cpi%202%5E2%5Ctimes%2010%5Ctimes%207.44%5C%5C%5CRightarrow%20F%3D387999.259089%5C%20N)
The weight of the column of methane is 387999.259089 N
The pressure at a depth is given by
![P=P_0+\rho gh\\\Rightarrow P=8.043\times 10^4\times +415\times 7.44\times 10\\\Rightarrow P=2.48335668\times 10^9\ Pa](https://tex.z-dn.net/?f=P%3DP_0%2B%5Crho%20gh%5C%5C%5CRightarrow%20P%3D8.043%5Ctimes%2010%5E4%5Ctimes%20%2B415%5Ctimes%207.44%5Ctimes%2010%5C%5C%5CRightarrow%20P%3D2.48335668%5Ctimes%2010%5E9%5C%20Pa)
The pressure at the depth is ![2.48335668\times 10^9\ Pa](https://tex.z-dn.net/?f=2.48335668%5Ctimes%2010%5E9%5C%20Pa)
Answer:
Half of the moon or 50%
Explanation:
The first quarter phase of the moon is when the Sun illuminates the half part of the moon while the other half rest in darkness.
In the first quarter phase of the moon, the illumination on the moon surface keeps on rising.
When the new moon appear, there is no illumination while the illumination is 100 % on Full moon.
For the phases of crescent moon, it ranges from 0 - 50% and then for gibbous phase, the range of the illumination is from 50-100%from new
Answer:
The magnitude of the centripetal acceleration during the turn is ![a=12.04\ m/s^2.](https://tex.z-dn.net/?f=a%3D12.04%5C%20m%2Fs%5E2.)
Explanation:
Given :
Speed to the airplane in circular path , v = 115 m/s.
Time taken by plane to turn , t= 15 s.
Also , the plane turns from east to south i.e. quarter of a circle .
Therefore, time taken to complete whole circle is , ![T=t\times 4=60\ s.](https://tex.z-dn.net/?f=T%3Dt%5Ctimes%204%3D60%5C%20s.)
Now , Velocity ,
![v=\dfrac{2\pi r}{T}\\\\115=\dfrac{2\times 3.14\times r}{60}\\\\r=1098.73\ m.](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B2%5Cpi%20r%7D%7BT%7D%5C%5C%5C%5C115%3D%5Cdfrac%7B2%5Ctimes%203.14%5Ctimes%20r%7D%7B60%7D%5C%5C%5C%5Cr%3D1098.73%5C%20m.)
Also , we know :
Centripetal acceleration ,
![a=\dfrac{v^2}{r}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2%7D%7Br%7D)
Putting all values we get :
![a=12.04\ m/s^2.](https://tex.z-dn.net/?f=a%3D12.04%5C%20m%2Fs%5E2.)
Hence , this is the required solution .
Both intake and exhaust valves are closed during most of the power stage in a four cycle gas engine.