7. PE=0.5×700n/m×0.9m^2
0.9^2=0.81m
0.5×700×0.81= 283.5J
8. 2000=0.5×(x)×1.5m^2
1.5^2= 0.25
0.25×0.5=0.125
2000=0.125 (x)
2000/0.125=x
x=16000 n/m
9. 4000=0.5 (375 n/m)×(x)^2
0.5×187.5 (x)
4000/187.5=21.3333333333
A magnet has a South Pole and a North Pole. South Pole and South Pole can't connect to her other, same as North and North. The same poles push each other away.
South Pole and North Pole connect.
Answer:
C. 110 m/s2
Explanation:
Force = Mass x Acceleration
Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:
Force/Mass = (Mass x Acceleration)/Mass
Acceleration = Force/Mass
Now we just have to plug in our values and calculate!
Acceleration = 48.4/0.44
Acceleration = 110m/s/s
It is option C. 110 m/s2
Hope this helped!
The new oscillation frequency of the pendulum clock is 1.14 rad/s.
The given parameters;
- <em>Mass of the pendulum, = M </em>
- <em>Length of the pendulum, = L</em>
- <em>Initial angular speed, </em>
<em> = 1 rad/s</em>
The moment of inertia of the rod about the end is given as;
The moment of inertia of the rod between the middle and the end is calculated as;
![I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}](https://tex.z-dn.net/?f=I_f%20%3D%20%5Cint%5Climits%5EL_%7BL%2F2%7D%20%7Br%5E2%5Cfrac%7BM%7D%7BL%7D%20%7D%20%5C%2C%20dr%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5Br%5E3%5D%5EL_%7BL%2F2%7D%20%3D%20%20%5Cfrac%7BM%7D%7B3L%7D%20%5BL%5E3%20-%20%5Cfrac%7BL%5E3%7D%7B8%7D%20%5D%20%3D%20%5Cfrac%7BM%7D%7B3L%7D%20%5B%5Cfrac%7B7L%5E3%7D%7B8%7D%20%5D%3D%20%5Cfrac%7B7ML%5E2%7D%7B24%7D)
Apply the principle of conservation of angular momentum as shown below;
Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.
Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129
Answer:
Explanation:
70.0(5.00)(9.81) = 3,433.5 = 3430 N
