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3 years ago

if you drop a stone from height of 2.5m. what is the speed of the stone right before it hits the ground?

1 answer:

6 0

Since the stone was dropped from height, its initial velocity = 0 m/s

Using v² = u² + 2gs.

Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m

v² = u² + 2gs

v² = 0² + 2*10*2.5

v² = 0 + 50

v² = 50

v = √50

v ≈ 7.07 m/s

Hence velocity just before hitting the ground is ≈ 7.07 m/s

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The magnitude of the magnetic field halfway between the wires is 3.0 x 10⁻⁵ T.

Explanation:

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$B = \frac{\mu _oI_1}{2 \pi r} + \frac{\mu_oI_2}{2\pi r}$

where;

B is magnitude of the magnetic field halfway between the wires

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I₂ is current the second wire

μ₀ is permeability of free space

r is distance half way between the wires

$B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r} (I_1 +I_2)\\\\B = \frac{4\pi *10^{-7} }{2\pi *0.2} (10 +20) = 3.0 *10^{-5}\ T$

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3 years ago