If we have I= 7.5 A:
I think my solution might just help you answer the problem on your own:
You have the formulas correct, watch your signs and BRACKETS.
B = μ0/(2π) (Current) / (Perpendicular distance)
Since μ0=4π E -7 Tm/A, we have:
B1 = (4πE-7 Tm/A)(7.5 A)/[2π (0.030 m)] = 5E-5 T
B2 = (4πE-7 Tm/A)(-7.5 A)/[2π (0.150 m)] = -1E-1 T
So BA = B1 + B2 = ?
(It looks like you just left out the square brackets, hence multiplying Pi and 0.03 and 0.15 instead of dividing them.)
<span>For the point B, the two distances are -0.060 m and +0.060 m. Be careful with the signs. Unlike point A, the two components will have the same sign.</span>
If a tennis player does not swing through, meaning they stop swinging the moment they make contact with the ball, they would lose the majority of their power. <em>The momentum that they had built up during the swing is lost the moment they stop swinging</em>, meaning that the ball is hit with a low amount of power.
<em>If the tennis player swings through the whole time they hit the ball, then they keep their momentum as they hit the ball.</em> There is a much higher power level when swinging through than if you were to stop your swing when you hit the ball.
Answer:
Magnification = 1
Explanation:
given data
radius of curvature r = - 0.983 m
image distance u = - 0.155
solution
we get here first focal length that is
Focal length, f = R/2 ...................1
f = -0.4915 m
we use here formula that is
.................2
put here value and we get
<h3>v = 0.155 m </h3>
so
Magnification will be here as
m =
m =
<h3>m = 1</h3>
Answer:
stone A is diamond.
Explanation:
given,
Volume of the two stone = 0.15 cm³
Mass of stone A = 0.52 g
Mass of stone B = 0.42 g
Density of the diamond = 3.5 g/cm³
So, to find which stone is gold we have to calculate the density of both the stone.
We know,
density of stone A
density of stone B.
Hence, the density of the stone A is the equal to Diamond then stone A is diamond.