Respuesta :
Answer:
The ball and the strings are shown in the attached figure
Balancing the forces in Y direction we have
[tex]35sin(\theta )=1.34\times 9.81+T_{2}sin(\theta )\\\\sin(\theta )=\frac{0.85}{1.7}=0.5\\\\\therefore T_{2}=\frac{35sin(\theta )-13.14N}{sin(\theta )}\\\\T_{2}=8.72N[/tex]
The magnitude of net force on ball in
1) Y direction is zero since there is no acceleration of ball along y direction.
2)In x direction is given by [tex]35cos(30)+T_{2}cos(30)=37.88N[/tex]
c)
The force in the x direction provides centripetal acceleration for the ball thus we have
[tex]35cos(30)+T_{2}cos(30)=37.88N\\\\37.88=\frac{mv^{2}}{r}\\\\\therefore v=\sqrt{\frac{37.88\times r}{m}}\\\\r=1.70cos(30)=1.47m\\\\\therefore v=\sqrt{\frac{37.88\times 1.47}{1.34}}=6.44m/s\\\\[/tex]
a) [tex]\rm T_2 = 8.72\;N[/tex]
b) Y direction is zero since there is no acceleration of ball along y direction.
  In x direction is given by
  [tex]\rm 35cos30 + 8.72cos30 = 37.88\;N[/tex]
c) [tex]\rm v = 6.44\; m/sec[/tex]
Given :
Mass of ball = 1.34 Kg
String length = 1.7 m
d = 1.7 m
Tension = 35 N
Solution :
a)
[tex]\rm sin\theta= \dfrac{0.85}{1.7}=0.5[/tex]
[tex]\theta = 30^\circ[/tex]
[tex]\rm T_2 = \dfrac{35sin\theta-13.14}{sin\theta}[/tex]
[tex]\rm T_2 = 8.72\;N[/tex]
b) Y direction is zero since there is no acceleration of ball along y direction.
  In x direction is given by
  [tex]\rm 35cos30 + 8.72cos30 = 37.88\;N[/tex]
c) We know that the net force in x direction is -
[tex]\rm 37.88 = \dfrac{mv^2}{r}[/tex]
[tex]\rm v= \sqrt{\dfrac{37.88\times r}{m}}[/tex] Â ---- (1)
Now,
[tex]\rm r = 1.7cos(30)= 1.47\; m[/tex]
Put the value of r and m in equation (1) we get,
[tex]\rm v= \sqrt{\dfrac{37.88\times 1.47}{1.34}}[/tex]
[tex]\rm v = 6.44\; m/sec[/tex]
For more information, refer the link given below
https://brainly.com/question/17447083?referrer=searchResults
