Answer:
The velocity of the penny at 25th second is, v = 245 m/s
Explanation:
Given,
The time taken by the penny to hit the ground is, t = 30 s
The initial velocity of the penny, u = 0 m/s
The velocity of the penny at 25th second, v = ?
Using the II equation of motion, the displacement is given by the formula
<em> S = ut + ½ gt²</em>
= 0 + ½ x 9.8 ( 25²)
= 3062.5 m
Using the III equations of motion
<em>v² = u² + 2gs</em>
Substituting the given values,
v² = 0 + 2 x 9.8 x 3062.5
= 60025
v = 245 m/s
Hence, the velocity of the penny at 25th second is, v = 245 m/s
Answer:
100.008 km/h
Explanation:
multiply the speed value by 3.6
Answer:
L = 16 [in]
Explanation:
In order to find the arch we must first find the angle that forms the circular sector, using the following formula:
![A = \frac{\alpha*r^{2} }{2} \\where:\\A= area = 64[in^{2}]\\r = radius = 8[in]\\\\\alpha = angle[rad]therefore\\\alpha =\frac{2*A}{r^{2} } \\\alpha =\frac{2*64}{8^{2} } \\\alpha =2[rad]](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B%5Calpha%2Ar%5E%7B2%7D%20%7D%7B2%7D%20%5C%5Cwhere%3A%5C%5CA%3D%20area%20%3D%2064%5Bin%5E%7B2%7D%5D%5C%5Cr%20%3D%20radius%20%3D%208%5Bin%5D%5C%5C%5C%5C%5Calpha%20%3D%20angle%5Brad%5Dtherefore%5C%5C%5Calpha%20%3D%5Cfrac%7B2%2AA%7D%7Br%5E%7B2%7D%20%7D%20%5C%5C%5Calpha%20%3D%5Cfrac%7B2%2A64%7D%7B8%5E%7B2%7D%20%7D%20%5C%5C%5Calpha%20%3D2%5Brad%5D)
Now using the following equation we can calculate the arc length
![L=\alpha *r\\L=2*8\\L=16[in]](https://tex.z-dn.net/?f=L%3D%5Calpha%20%2Ar%5C%5CL%3D2%2A8%5C%5CL%3D16%5Bin%5D)
It's called induced charge separation I believe.
Answer:
808.25 seconds
Explanation:
From the relation, ![Velocity = \frac{Length}{time}](https://tex.z-dn.net/?f=Velocity%20%3D%20%5Cfrac%7BLength%7D%7Btime%7D)
If t = time
v = velocity
l = length
![t = \frac{l}{v} \\](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7Bl%7D%7Bv%7D%20%5C%5C)
The current density, J, of a charge is given by:
![J = nqv..........................(1)\\J = \frac{I}{A} ................(2)](https://tex.z-dn.net/?f=J%20%3D%20nqv..........................%281%29%5C%5CJ%20%3D%20%5Cfrac%7BI%7D%7BA%7D%20................%282%29)
Where number of charge carriers per unit volume n = 8.49 * 10²⁸
Charge of the electron, q = 1.6 * 10⁻¹⁹
Cross sectional area of the wire, A = 0.21 cm² = 0.000021 m²
Equating (1) and (2)
![nqv = \frac{I}{A} \\v = \frac{I}{nqA} \\](https://tex.z-dn.net/?f=nqv%20%3D%20%5Cfrac%7BI%7D%7BA%7D%20%5C%5Cv%20%3D%20%5Cfrac%7BI%7D%7BnqA%7D%20%5C%5C)
Since t = l/v.......(3)
Put the formula for v into equation (3)
![t = \frac{nqAl}{I} \\t = \frac{8.49*10^{28}*1.6 *10^{-19}* 2.1 * 10^{-5} * 0.85}{300}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BnqAl%7D%7BI%7D%20%5C%5Ct%20%3D%20%5Cfrac%7B8.49%2A10%5E%7B28%7D%2A1.6%20%2A10%5E%7B-19%7D%2A%202.1%20%2A%2010%5E%7B-5%7D%20%2A%200.85%7D%7B300%7D)
t = 808.25 seconds