What is the question? if you are looking for the initial velocity of the projectile, then V escape of Earth = 11.2 km / s. divided by three
Vi = 3.73 km/s
Because inside the gun it has dynamite
Answer:
1.78 rad/s
1.70344 rad/s
Explanation:
v = Velocity = 0.8 m/s
m = Mass of person = 60 kg
= Distance between center of mass of person and pole = 0.45 m
= New distance between center of mass of person and pole = 0.46 m
I = Moment of inertia
Angular speed is given by
![\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s](https://tex.z-dn.net/?f=%5Comega_1%3D%5Cdfrac%7Bv%7D%7Br_1%7D%5C%5C%5CRightarrow%20%5Comega_1%3D%5Cdfrac%7B0.8%7D%7B0.45%7D%5C%5C%5CRightarrow%20%5Comega_1%3D1.78%5C%20rad%2Fs)
The angular speed is 1.78 rad/s
In this system the angular momentum is conserved
![L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s](https://tex.z-dn.net/?f=L_1%3DL_2%5C%5C%5CRightarrow%20I_1%5Comega_1%3DI_2%5Comega_2%5C%5C%5CRightarrow%20mr_1%5E2%5Comega_1%3Dmr_2%5E2%5Comega_2%5C%5C%5CRightarrow%20%5Comega_2%3D%5Cdfrac%7Br_1%5E2%5Comega_1%7D%7Br_2%5E2%7D%5C%5C%5CRightarrow%20%5Comega_2%3D%5Cdfrac%7B0.45%5E2%5Ctimes%201.78%7D%7B0.46%5E2%7D%5C%5C%5CRightarrow%20%5Comega_2%3D1.70344%5C%20rad%2Fs)
The new angular speed is 1.70344 rad/s
Answer:
if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).
Answer:
a). 53.78 m/s
b) 52.38 m/s
c) -75.58 m
Explanation:
See attachment for calculation
In the c part, The negative distance is telling us that the project went below the lunch point.