Answer:
The magnitude of the magnetic field at the origin is 
.
Explanation:
Given :
50-A wire is in the x-z plane and is 5 m from the z axis.
Also , 40-A wire is in the y-z plane and is 4 m from the z axis.
Now , since both the wire are perpendicular to each other .
Therefore , magnetic field are also perpendicular to each other .
Magnetic field at origin due to wire 1 is :
Magnetic field at origin due to wire 2 is :
Now , therefore net magnetic field is :
Putting value of 
We get ,
Therefore, the magnitude of the magnetic field at the origin is .
The pressure of the atmosphere, when a barometer reads 780 mm Hg. Mercury which a density of 1.36 x 10^4 kg /m^3 is B 1.1 x 10^5 N/m^2
This problem can be solved using the formula below
P = dgh................. Equation 1
Where P = Pressure of the atmosphere, d = density of the mercury, h = height of the mercury, g = acceleration due to gravity.
From the question,
Given: d = 1.36×10⁴ kg/m³, h = 780 mm = 0.78 m,
Constant: g = 10 m/s²
Substitute these values into equation 1
P = (1.36×10⁴)(10)(0.78)
P = 10.608×10⁴ N/m²
P ≈ 1.1×10⁵ N/m²
Hence the right answer is B. 1.1×10⁵ N/m²
Learn more about Pressure here: brainly.com/question/23603188
First, let us derive our working equation. We all know that pressure is the force exerted on an area of space. In equation, that would be: P = F/A. From Newton's Law of Second Motion, force is equal to the product of mass and gravity: F = mg. So, we can substitute F to the first equation so that it becomes, P = mg/A. Now, pressure can also be determined as the force exerted by a fluid on an area. This fluid can be measure in terms of volume. Relating volume and mass, we use the parameter of density: ρ = m/V. Simplifying further in terms of height, Volume is the product of the cross-sectional area and the height. So, V = A*h. The working equation will then be derived to be:
P = ρgh
This type of pressure is called the hydrostatic pressure, the pressure exerted by the fluid over a known height. Next, we find the literature data of the density of seawater. From studies, seawater has a density ranging from 1,020 to 1,030 kg/m³. Let's just use 1,020 kg/m³. Substituting the values and making sure that the units are consistent:
P = (1,020 kg/m³)(9.81 m/s²)(11 km)*(1,000 m/1km)
P = 110,068,200 Pa or 110.07 MPa
2 is B. 3 is D. 4 is C. I think 5 is A. 6 is A. 7 is D. I think you are all correct. Good Luck!
