Answer:
Explanation:
<u>Given Data:</u>
Length = l = 820 mm = 0.82 m
Acceleration due to gravity = g = 9.8 ms⁻²
<u>Required:</u>
Frequency = f = ?
<u>Formula:</u>
<u>Solution:</u>
![\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%20%3D%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7Bg%7D%7Bl%7D%20%7D%20%5C%5C%5C%5CPut%5C%20the%5C%20givens%5C%5C%5C%5Cf%3D%5Cfrac%7B1%7D%7B2%20%5Cpi%7D%20%5Csqrt%7B%5Cfrac%7B9.8%7D%7B0.82%7D%20%7D%5C%5C%5C%5C%20f%20%3D%200.159%20%5Ctimes%20%5Csqrt%7B11.95%7D%20%5C%5C%5C%5Cf%3D0.159%20%5Ctimes%203.457%5C%5C%5C%5Cf%3D0.55%20%5C%20Hz%5C%5C%5C%5C%5Crule%5B225%5D%7B225%7D%7B2%7D)
Answer:
b. electric potential energy.
Explanation:
The energy required to move a charge against the electric field is known as the electric potential energy. As in above case positively charged body is exerting an electric field on the positive charge. As the same charges repel so the charge tend to move away. In order to push it towards the body we need a work done. As it is hard to push the positive charged particle towards the positive electric field. So in the cases like these particle occupies the electric potential energy.
1 watt = 1 joule per second = 1 newton meter per second = 1 kg m2 s-3
<span>Using conservation of energy and momentum you can solve this question. M_l = mass of linebacker
M_ h = mass of halfback
V_l = velocity of linebacker
V_h = velocity of halfback
So for conservation of momentum,
rho = mv
M_l x V_li + M_h x V_hi = M_l x V_lf + M_h x V_hf
For conservation of energy (kinetic)
E_k = 1/2mv^2/ 1/2mV_li^2 + 1/2mV_{hi}^2 = 1/2mV_{lf}^2 + 1/2mV_{hf}^2
Where i and h stand for initial and final values.
We are already told the masses, \[M_l = 110kg\] \[M_h = 85kg\] and the final velocities \[V_{fi} = 8.5ms^{-1}\] and \[V_{ih} = 7.2ms^{-1} </span>
