Mgh= 1/2 m v^2
gh= 1/2 v^2 9.8 * 8= 1/2 v^2
solve for v
V = sqrt(2gH)where H = 8 m.
Answer:the maximum Hall voltage across the strip= 0.00168 V.
Explanation:
The Hall Voltage is calculated using
Vh= B x v x w
Where
B is the magnitude of the magnetic field, 5.6 T
v is the speed/ velocity of the strip, = 25 cm/s to m/s becomes 25/100=0.25m/s
and w is the width of the strip= 1.2 mm to meters becomes 1.2 mm /1000= 0.0012m
Solving
Vh= 5.6T x 0.25m/s x 0.0012m
=0.00168T.m²/s
=0.00168Wb/s
=0.00168V
Therefore, the maximum Hall voltage across the strip=0.00168V
Answer:
the spring be displaced by 25.0 cm
Explanation:
The computation is shown below:
As we know that
F= -K × x
So,

Now

= -0.250m
= 25.0 cm
Hence, the spring be displaced by 25.0 cm
Answer:
1Mm
Explanation:
i think this one. so it is wrong then also dont be angry with me because iam trying to help u
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