A mass M = 4 kg attached to a string of length L = 2.0 m...

Question

Question

Answer 1

Answer #1

Tcono sino = 5 / r= lsino T sino a r = 2x Sin441 ora 1.39 m To= my T cono= my Tsino mue ус ол Tano = 2 v²rg Tamo v=J 1.39x9.8

Answer 2

Similar Homework Help Questions

A mass M = 4 kg attached to a string of length L = 2.0 m...

A mass M = 4 kg attached to a string of length L = 2.0 m swings in a horizontal circle with a speed V. The string maintains a constant angle theta θ= 46.5 degrees with the vertical line through the pivot point as it swings. What is the speed V required to make this motion possible?
A mass m = 4.700 kg is suspended from a string of length L = 1.270...

A mass m = 4.700 kg is suspended from a string of length L = 1.270 m. It revolves in a horizontal circle. The tangential speed of the mass is 2.243 m/s. What is the angle theta between the string and the vertical (in degrees)?
A mass m = 8.3 kg is suspended from a string of length L = 1.33...

A mass m = 8.3 kg is suspended from a string of length L = 1.33 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.27 m/s. What is the angle between the string and the vertical (in degrees)?

A mass m = 4.300 kg is suspended from a string of length L = 1.290...

A mass m = 4.300 kg is suspended from a string of length L = 1.290 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.743 m/s. What is the angle theta between the string and the vertical (in degrees)?
A mass m = 7.9 kg is suspended from a string of length L = 1.15...

A mass m = 7.9 kg is suspended from a string of length L = 1.15 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 2.80 m/s. What is the angle θ between the string and the vertical (in degrees)?
2. [2pt] A mass m = 9.100 kg is suspended from a string of length L...

2. [2pt] A mass m = 9.100 kg is suspended from a string of length L = 1.210 m. It revolves in a horizontal circle (see Figure). The tangential speed of the mass is 3.089 m/s. What is the angle theta between the string and the vertical (in degrees)? Answer: Submit All Answers

A mass m = 0.6kg is attached to a string of length L = 1.5m. The...

A mass m = 0.6kg is attached to a string of length L = 1.5m. The string is pulled in such a way that it makes an angle of 25 degrees with the vertical direction as shown in the figure below. The mass is released from rest. a. Find the speed of as the mass passes through the bottom of its trajectory. b. What is the tension in the string when the mass passes through the bottom of its trajectory?...
This mass (m) on a string (of length L) is moving in a horizontal circle. The...

This mass (m) on a string (of length L) is moving in a horizontal circle. The string makes an angle of 60 degrees with the vertical. (Express your answers in terms of m, L, g, and theta) Find the tension in the string. Find the speed of the mass.
A rock of mass m = 0.97 kg is attached to a massless string of length...

A rock of mass m = 0.97 kg is attached to a massless string of length L = 0.71 m. The rock is swung in a vertical circle faster and faster up to a speed of v = 2.4 m/s, at which time the string breaks. What is the magnitude of the tension, in newtons, at which the string breaks?

9) A ball of mass m is attached to the ceiling via a string of length...

9) A ball of mass m is attached to the ceiling via a string of length L. The ball is given a push to set it in motion. The ball then travels in a horizontal circle, such that the string makes and angle 0 with the vertical. It is observed that it takes time T for the ball to make one complete revolution. How can this experiment be used to measure the magnitude of the gravitational acceleration g. problem 9...