Given that:
k = 500 n/m,
work (W) = 704 J
spring extension (x) = ?
we know that,
Work = (1/2) k x²
704 = (1/2) × 500 × x²
x = 1.67 m
A spring stretched for 1.67 m distance.
Answer:
The other angle is 120°.
Explanation:
Given that,
Angle = 60
Speed = 5.0
We need to calculate the range
Using formula of range
...(I)
The range for the other angle is
....(II)
Here, distance and speed are same
On comparing both range
![\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}](https://tex.z-dn.net/?f=%20%5Cdfrac%7Bv%5E2%5Csin%282%5Ctheta%29%7D%7Bg%7D%3D%5Cdfrac%7Bv%5E2%5Csin%282%28%5Calpha-%5Ctheta%29%29%7D%7Bg%7D)
![\sin(2\theta)=\sin(2\times(\alpha-\theta))](https://tex.z-dn.net/?f=%5Csin%282%5Ctheta%29%3D%5Csin%282%5Ctimes%28%5Calpha-%5Ctheta%29%29)
![\sin120=\sin2(\alpha-60)](https://tex.z-dn.net/?f=%5Csin120%3D%5Csin2%28%5Calpha-60%29)
![120=2\alpha-120](https://tex.z-dn.net/?f=120%3D2%5Calpha-120)
![\alpha=\dfrac{120+120}{2}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B120%2B120%7D%7B2%7D)
![\alpha=120^{\circ}](https://tex.z-dn.net/?f=%5Calpha%3D120%5E%7B%5Ccirc%7D)
Hence, The other angle is 120°
The centripetal acceleration is given by
![a_c = \frac{v^2}{r}](https://tex.z-dn.net/?f=a_c%20%3D%20%20%5Cfrac%7Bv%5E2%7D%7Br%7D%20)
where v is the tangential speed and r the radius of the circular orbit.
For the car in this problem,
![a_c = 15.625 m/s^2](https://tex.z-dn.net/?f=a_c%20%3D%2015.625%20m%2Fs%5E2)
and r=40 m, so we can re-arrange the previous equation to find the velocity of the car: