Answer:
they are happy cause he jumped off a bridge
Explanation:
Answer:
![\boxed {\boxed {\sf 56.25 \ Joules}}](https://tex.z-dn.net/?f=%5Cboxed%20%7B%5Cboxed%20%7B%5Csf%2056.25%20%5C%20Joules%7D%7D)
Explanation:
Kinetic energy can be found using the following formula:
![E_k=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
where <em>m </em>is the mass and <em>v</em> is the velocity.
The mass of the ball is 0.5 kilograms and the velocity is 15 meters per second
![m= 0.5 \ kg \\v= 15 \ m/s \\](https://tex.z-dn.net/?f=m%3D%200.5%20%5C%20kg%20%5C%5Cv%3D%2015%20%5C%20m%2Fs%20%5C%5C)
Substitute the values into the formula.
![E_k=\frac{1}{2} (0.5 \ kg ) * (15 \ m/s)^2](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7D%20%280.5%20%5C%20kg%20%29%20%2A%20%2815%20%20%5C%20m%2Fs%29%5E2)
First, evaluate the exponent.
- (15 m/s)²= (15 m/s) * (15 m/s) = 225 m²/s²
![E_k=\frac{1}{2} (0.5 \ kg ) *(225 \ m^2/s^2)](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7D%20%280.5%20%5C%20kg%20%29%20%2A%28225%20%5C%20m%5E2%2Fs%5E2%29)
Multiply 0.5 kg by 225 m²/s²
![E_k=\frac{1}{2}(112.5 \ kg* m^2/s^2)](https://tex.z-dn.net/?f=E_k%3D%5Cfrac%7B1%7D%7B2%7D%28112.5%20%5C%20kg%2A%20m%5E2%2Fs%5E2%29)
Multiply 112 kg*m²/s² by 1/2, or divide by 2.
![E_k=56.25 \ kg*m^2/s^2](https://tex.z-dn.net/?f=E_k%3D56.25%20%5C%20kg%2Am%5E2%2Fs%5E2)
- 1 kg*m²/s² is equal to 1 Joule
- Therefore, our answer of 56.25 kg*m²/s² is equal to 56.25 Joules.
![E_k= 56.25 \ J](https://tex.z-dn.net/?f=E_k%3D%2056.25%20%5C%20J)
The kinetic energy of the ball is <u>56.25 Joules</u>
Answer: she did no work while she posed because she did not move the torch or the book
Explanation: just did it on edge
Answer:
![F_0 = 393 N](https://tex.z-dn.net/?f=F_0%20%3D%20393%20N)
Explanation:
As we know that amplitude of forced oscillation is given as
![A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7BF_0%7D%7B%20m%28%5Comega%5E2%20-%20%5Comega_0%5E2%29%7D)
here we know that natural frequency of the oscillation is given as
![\omega_0 = \sqrt{\frac{k}{m}}](https://tex.z-dn.net/?f=%5Comega_0%20%3D%20%5Csqrt%7B%5Cfrac%7Bk%7D%7Bm%7D%7D)
here mass of the object is given as
![m = \frac{W}{g}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7BW%7D%7Bg%7D)
![\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}](https://tex.z-dn.net/?f=%5Comega_0%20%3D%20%5Csqrt%7B%5Cfrac%7B220%7D%7B%5Cfrac%7B30%7D%7B9.81%7D%7D%7D)
![\omega_0 = 8.48 rad/s](https://tex.z-dn.net/?f=%5Comega_0%20%3D%208.48%20rad%2Fs)
angular frequency of applied force is given as
![\omega = 2\pi f](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%20f)
![\omega = 2\pi(10.5) = 65.97 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%202%5Cpi%2810.5%29%20%3D%2065.97%20rad%2Fs)
now we have
![0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}](https://tex.z-dn.net/?f=0.03%20%3D%20%5Cfrac%7BF_0%7D%7B3.06%2865.97%5E2%20-%208.48%5E2%29%7D)
![F_0 = 393 N](https://tex.z-dn.net/?f=F_0%20%3D%20393%20N)
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