Answer:
The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F is force and is a vector because the force acts in a direction. q is the charge of the particle. v is velocity and is a vector because the particle is moving in some direction. B is the magnetic flux density.
We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd. Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore
∮B dl=μ0 I
B ∮dl=μ0 I
B 2πr=μ0 I
B=μ02πr Id=μ0/4π I dl×rr3
Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get
dl×r=−z dlj^+rsinθk^
and therefore the magnitude of dB is equal to
dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3
Thus, magnetic field is depending on r,θ,z.
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To answer this question, first we take note that the maximum height that can be reached by an object thrown straight up at a certain speed is calculated through the equation,
Hmax = v²sin²θ/2g
where v is the velocity, θ is the angle (in this case, 90°) and g is the gravitational constant. Since all are known except for v, we can then solve for v whichi s the initial velocity of the projectile.
Once we have the value of v, we multiply this by the total time traveled by the projectile to solve for the value of the range (that is the total horizontal distance).
Answer:
Target ceiling. the upper limit of your physical activity. Target fitness zone. Above the threshold of training and below the target ceiling.
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Explanation:
The answer is D light rays shine on an object which then reflects back to our retina
When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:
H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s
t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s
Total time = 0.125 s + 4.495 s = 4.62 seconds
