$Formula\ for\ period:\\\ T=2 \pi \sqrt{\frac{L}{g}}\\\ g-gravity=9,8 \frac{m}{s^2} ,\ L-pendulum \ length \\\\ \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\\\ \frac{T^2}{2 \pi } = \frac{L}{g} \\\\\ \frac{T^2}{2 \pi }*g=L\\\\ L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m$ $T=2 \pi \sqrt{\frac{L}{g}} \\
 \frac{T}{2 \pi } = \sqrt{ \frac{L}{g} }|square\\
 \frac{T^2}{2 \pi } = \frac{L}{g} \\
 \frac{T^2}{2 \pi }*g=L\\
L= \frac{2^2}{2*3,14 }*9,8= \frac{39,2}{6,28} =6,24m
$

7 0

3 years ago

Current what can do work in an electric device

Serggg [28]

It could help transport important information

5 0

3 years ago

A 0.810 kg ball falls 2.5m. How much work does the force of gravity do on the ball?

lisabon 2012 [21]

Answer:

W = 19.845 J

Explanation:

Work is defined as W = Fdcos $\theta$ , where F is the force exerted and d is the distance. Because the direction the ball is falling is the same direction as the force itself, $\theta$ = 0 deg, and since cos(0) = 1, this equation is equivalent to W = Fd. In this case, the force exerted is the weight force, which is equivalent to m * g. Substituting you get:

W = mgd = 0.810 kg * 9.8 m/s^2 * 2.5m

W = 19.845 J

6 0

2 years ago