Answer:
The answer to the question is
The object would fall 57.625 m in the first 5 seconds
Explanation:
To solve the question, we note that
the height of fall = 490 ft = 149.352 m
Time to touch the ground = 7 seconds
We are required to find out how far the object falls in the first 5 seconds
We apply the relation
S = u·t + 0.5×g·t ² = We then have
149.352 = U×7+0.5*9.81*49 From where u = -13 m/s
Therefore to find how far it falls in the first 5 seconds, we have
-13*5 + 0.5*9.81*25 = 57.625 m
14 ms is required to reach the potential of 1500 V.
<u>Explanation:</u>
The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

As two different current is passing at two different times, the net charge will be the different in current. So,

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

Here
, q is the charge and R is the radius. As
and R =17 cm = 0.17 m, then the voltage will be

The time is required to find to reach the voltage of 1500 V, so


So, 14 ms is required to reach the potential of 1500 V.
The fast sports car does more damage then the slow semi truck
Answer:
v = 5.24[m/s]
Explanation:
Este problema se puede resolver por medio del principio de la conservación de la energía, donde la energía potencial es igual a la energía cinética. Es decir a medida que el carrito desciende su energía potencial disminuye, pero su energía cinética aumenta.

Donde:

Ahora reemplazando:
![\frac{1}{2} *m*v^{2}=m*g*h\\\\0.5*v^{2}=9.81*1.4\\v=\sqrt{\frac{9.81*1.4}{0.5} } \\\\v=5.24[m/s]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%3Dm%2Ag%2Ah%5C%5C%5C%5C0.5%2Av%5E%7B2%7D%3D9.81%2A1.4%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B9.81%2A1.4%7D%7B0.5%7D%20%7D%20%20%20%5C%5C%5C%5Cv%3D5.24%5Bm%2Fs%5D)